$G$ cocycle split and trivialized to a coboundary in $J$, given a group homomorphism $J \overset{r}{\rightarrow} G$
A small remark first: I believe that the denominator in the exponent should be 2, and not $2^2$. In any case, write $Q_8=\langle x,y|x^2=y^2, xyx^{-1}=y^{-1},y^4=1\rangle$ so that each element in the group we can write uniquely as $y^ax^b$ with $a\in\{0,1,2,3\}$ and $b\in\{0,1\}$. By manipulating a bit the Lyndon Hochschild Serre spectral sequence, one gets the function $$\beta(y^ax^b,y^cx^d) = i^{bc}.$$ This function satisfies the desired property. Let me explain how to receive this $\beta$: Let us write $Q_8=G$, the normal subgroup as $N$ and the quotient group as $Q$. By choosing a set-theoretic lifting $s:Q\to G$, we can write each element of $G$ uniquely as a product $ns(q)$ for some $n\in N$ and $q\in Q$. Let us identify in this way $G$ with $N\times Q$ as a set. The key point here is that in the Lyndon Hochschild Serre spectral sequence you have a homomorphism $d_2:H^1(Q,H^1(N,U(1)))\to H^3(Q,U(1))$. Every element in the image of this $d_2$ will be in the kernel of the inflation map. In our case the group $H^1(Q,H^1(N,U(1)))$ is cyclic of order 2. It is generated by the map $f$ which sends the generator of $Q$ to the generator of the cyclic group of order 4, $H^1(N,U(1))$ (which sends a generator of $N$ to $i$). The idea now is that you try to cook up a two cocycle out of this element in the following way: $$\beta((n_1,q_1),(n_2,q_2)) = f(q_1^{-1})(n_2).$$ However, this $\beta$ might not be a two cocycle. The obstruction for this is of course $\partial \beta$. By a direct calculation, this is a three cocycle inflated from the quotient group $Q$, and this is exactly $d_2(f)$.
Here is an answer for question 2 on which homomorphisms $J\xrightarrow{r} \mathbb{Z}_2$ will trivialize $\omega_3^G$.
Your cocycle takes values in the two-element subgroup $\{1,-1\}\subset U(1)$ (aka $\{0,\frac{1}{2}\}\subset \mathbb{R}/\mathbb{Z}$). Since all two-element groups are isomorphic, you could of course identify this subgroup with $\mathbb{F}_2=\{0,1\}$ under addition. But why would you help? Well, $\mathbb{F}_2=\{0,1\}$ also has a multiplication (that's why I'm using a different letter for it), and so you can ask whether your cocycle factors as a cup product of two smaller cocycles. In this case, we'll see that it does! Let's see what happens in this case (this will also illustrate what a cup product is, if you're not familiar with this).
To start, we need to rewrite your formula $$\omega_3^{G}(g_1,g_2,g_3) = \exp \left( \frac{2 \pi i}{2^{2}} \; g_1(g_2 +g_3 -[(g_2 +g_3)\mod 2]) \right)\in U(1)$$ to take values in $\mathbb{F}_2=\{0,1\}$. Note that the inputs $g_i$ in your formula live in $\mathbb{Z}_2=\{0,1\}$. What we get (for now) is $$\omega_3(g_1,g_2,g_3)=g_1\cdot \frac{g_2 +g_3 -[(g_2 +g_3)\mod 2]}{2}\in \{0,1\}=\mathbb{F}_2.$$
The property we want to highlight is that this is a product of two simpler functions:
$$\omega_3(g_1,g_2,g_3)=\alpha(g_1)\cdot \beta(g_2,g_3)$$ where $$\alpha(g_1)=g_1\quad\text{ and }\quad\beta(g_2,g_3)=\frac{g_2 +g_3 -[(g_2 +g_3)\mod 2]}{2}.$$
What does this tell us? Well mathematically, we would write $r^*(\omega_3^G)\in H^2(J)$ for what you are calling $\omega_3^J$. And if $w_3\in H^3(\mathbb{Z}_2;\mathbb{F}_2)$ is the cohomology class represented by $\omega_3^G$, then $r^*(w_3)\in H^3(J;\mathbb{F}_2)$ is represented by $\omega_3^J$: so your question about trivializing the cocycle is whether $r^*(w_3)\overset{?}{=}0\in H^3(J;\mathbb{F}_2)$.
Note that $\alpha$ and $\beta$ are both cocycles, so they represent cohomology classes $a\in H^1(\mathbb{Z}_2;\mathbb{F}_2)$ and $b\in H^2(\mathbb{Z}_2;\mathbb{F}_2)$. The factorization $$\omega_3(g_1,g_2,g_3)=\alpha(g_1)\cdot \beta(g_2,g_3)$$ then tells us that $w_3$ is the cup product of the cohomology classes $a$ and $b$: we have $w_3=a\cdot b$ (see my answer here for details).
Cup product is an operation on cohomology that takes $x\in H^i$ and $y\in H^j$ and combines them to give $x\cdot y\in H^{i+j}$. It satisfies $y\cdot x = \pm x\cdot y$ where $\pm=(-1)^{ij}$.
The reason this is useful is that cup products are always preserved by $r^*$: there is a universal formula $$r^*(x\cdot y)=r^*(x)\cdot r^*(y)\quad\text{in cohomology}.$$ This means that ONE way you could hope to show that $r^*(w_3)=0$ would be to show that either $r^*(a)=0$ or $r^*(b)=0$. Actually in 1st cohomology $r^*(a)$ will never be 0 unless the map $J\xrightarrow{r}\mathbb{Z}_2$ is trivial (an uninteresting case, as you discuss in comments), but you could hope that $r^*(b)=0$. [Side note: in your particular example in Q1, I checked and $r^*(b)\neq 0$.]
But actually there is a much better way to look at this. It comes from the fact that you wrote your cocycle in a way that hides its symmetry. I claim that if you work out the values of your cocycle $$\omega_3(g_1,g_2,g_3)=g_1\cdot \frac{g_2 +g_3 -[(g_2 +g_3)\mod 2]}{2}\in \{0,1\}=\mathbb{F}_2$$ on the 8 different possible inputs, you'll see that it can instead be written as $$\omega_3(g_1,g_2,g_3)=\begin{cases}1&\text{ if $g_1=1$, $g_2=1$, and $g_3=1$}\\0&\text{else}\end{cases}\in \{0,1\}=\mathbb{F}_2.$$ In other words, $$\omega_3(g_1,g_2,g_3)=g_1\cdot g_2\cdot g_3=\alpha(g_1)\cdot \alpha(g_2)\cdot \alpha(g_3).$$ This means that $w_3=a\cdot a \cdot a=a^3$. So not only can your cocycle be written in a simpler form, it actually is the cup product of a single class in 1st cohomology with itself three times.
Let us write $R\in H^1(J;\mathbb{F}_2)$ for $R=r^*(a)$. Then we have $r^*(w_3)=r^*(a)^3=R^3$, so the question of whether $r^*(w3)=0$ is simply the question of whether $R\in H^1(J;\mathbb{F}_2)$ satisfies $$R^3=0\in H^3(J;\mathbb{F}_2).$$
Moreover, an element $R\in H^1(J;\mathbb{F}_2)$ is equivalent to a homomorphism $J\xrightarrow{r}\mathbb{Z}_2$ (that's why I used the same letter). So for a given $J$, your question is equivalent to finding all $R\in H^1(J;\mathbb{F}_2)$ satisfying $R^3=0$. Now, that's not going to necessarily be easy! But it is more likely to be something you can look up somehow. (And in some cases, you may be able to show that $R^2=0$, which would be enough.)
One last comment: you might be concerned that I was working with $\mathbb{F}_2$ coefficients here, whereas you had general $U(1)$ coefficients. Here is an argument why this is not so bad. Say we're given some class $z\in H^1$. If we take $x=z$ and $y=z$ in the formula $y\cdot x = \pm x\cdot y$ from above, it tells us $z\cdot z = - z\cdot z$. In other words, we always have $z^2+z^2=0\in H^2$. This means that if you have coefficients in $\mathbb{Q}$ or $\mathbb{Z}_3\subset U(1)$ or anywhere you can divide by 2, then $z^2$ would ALWAYS be 0 (and thus $z^3$ would be as well). This isn't quite a proof that you can always stick to $\mathbb{F}_2$ coefficients (for example, you might need to worry about $\mathbb{Z}/4\mathbb{Z}\subset U(1)$ as well) but as a general rule it argues that you can probably ignore factors other than $2$.
EDIT: Here is a concrete example of how to apply this to $Q_8$. According to (eq uation 2.4.2) of this survey by Benson, the cohomology ring of $Q_8$ is $$H^*(Q_8;\mathbb{F}_2)\cong \mathbb{F}_2[x,y,z]/(\,x^2+xy+y^2,\,x^2y+xy^2\,)$$ where $x\in H^1$, $y\in H^1$, and $z\in H^4$.
Therefore $H^3(Q_8;\mathbb{F}_2)$ is spanned by the four elements $x^3$, $x^2y$, $xy^2$, and $y^3$ subject to the three relations $x^3+x^2y+xy^2=0$, $x^2y+xy^2+y^3=0$, and $x^2y+xy^2=0$. (The first two are $x(x^2+xy+y^2)$ and $y(x^2+xy+y^2)$.) Subtracting the last relation from the first two gives a basis of relations as $x^3=0$, $y^3=0$, and $x^2y+xy^2=0$.
You can now read off easily exactly which $R\in H^1(Q_8;\mathbb{F}_2)$ have $R^3=0$. Namely, if $R=x$ we have $R^3 = x^3 = 0$; if $R=y$ we have $R^3 = y^3 = 0$; and if $R=x+y$ we have $R^3 = (x+y)^3 = x^3 +3x^2y+3xy^2+y^3 = x^3 + (x^2y+xy^2) + y^3 = 0$. Therefore we conclude that every homomorphism $Q_8\to \mathbb{F}_2$ splits your cocycle!
You can do this for lots of other groups, if you can find where someone has already computed the $\mathbb{F}_2$-cohomology ring. For example, (eq. 2.8.1) of that same survey says that the "semidihedral group $SD_{2^n}$ of order $2^n$", whatever that is, has (for $n\geq 4$) the cohomology ring $$H^*(SD_{2^n};\mathbb{F}_2)\cong \mathbb{F}_2[x,y,z,w]/(\,xy,\,y^3,\,yz,\,z^2+x^2w\,)$$ where $x\in H^1$, $y\in H^1$, $z\in H^3$, and $w\in H^4$. We find that $H^3(SD_{2^n};\mathbb{F}_2)$ is spanned by $x^3,x^2y,xy^2,y^3,z$ but with relations $x^2y=0$, $xy^2=0$, and $y^3=0$. Therefore we conclude that $R=x$ has $R^3=x^3\neq 0$ and $R=x+y$ has $R^3 = (x+y)^3 \equiv x^3 \neq 0$ but $R=y$ has $R^3=y^3=0$. So only one of the three nontrivial homomorphisms $SD_{2^n}\xrightarrow{r}\mathbb{Z}_2$ will split your cocycle.
Finally, you only have to look at groups $J$ whose order is a power of 2. This follows from a theorem of Cartan-Eilenberg, although it's a bit tricky to explain how (see Theorem 4.1 of these lectures by Adem). In any case, here is a precise statement you can use:
Theorem: Fix a homomorphism $J\xrightarrow{r}\mathbb{Z}_2$. For any subgroup $H<J$ whose order is a power of 2, you can consider the restriction $H\xrightarrow{r|_H}\mathbb{Z}_2$. Then the homomorphism $J\xrightarrow{r}\mathbb{Z}_2$ splits your cocycle $\iff$ for every subgroup $H<J$ whose order is a power of 2, the homomorphism $H\xrightarrow{r|_H}\mathbb{Z}_2$ splits your cocycle.