Converse to Stokes' Theorem

Yes $\omega$ and $\eta$ have to be linear.

We will need the following generalization of Minkowski's theorem on the existence of polyhedra with prescribed surface normals; see Theorem 1, p. 475 in "Gaussian images of surfaces and ellipticity of surface area functionals" by D. Burago and S. Ivanov. Note that the Grassman space $G(k,n)$ of $k$-planes in $\mathbb R^n$ can be thought as a set of unit unit simple $k$-vectors of $\mathbb R^n$.

Let $\mu$ be a measure on $G(n,k)$ and $\varepsilon>0$. Assume $$\int\limits_{G(k,n)}x\cdot\mu=0.$$ Then there is a closed $k$-dimensional submanifold $S$ in $\mathbb R^n$ such that the pushforward measure of $\mathrm{\rm vol}_k$ along the Gauss map in $G(k,n)$ is $\varepsilon$-close to $\mu$.

Now let us start. First we prove that $\eta$ is linear. It will be sufficient for us that $\int_N\eta=0$ for any closed $(k+1)$-submanifold $N$.

It is straightforward to reduce the question to the Euclidean space with parallel $k$-density. Once it is done, note that linearity of $\eta$ is equivalent to the fact that if $s_1,s_2,\dots,s_n$ be a collection of simple $(k+1)$-vectors such that $$s_1+s_2+\dots+s_n=0$$ then $$\eta(s_1)+\eta(s_2)+\dots+\eta(s_n)=0.$$

To show that the later property holds for $\omega$, we use Burago--Ivanov theorem for the measure $\mu$ on the Grassman with support at $\{\tfrac{s_i}{|s_i|}\}$ and such that $\mu(\tfrac{s_i}{|s_i|})=|s_i|$ for each $i$.

Now we will use linarity of $\eta$ to show that $\omega$ is also linear.

Note that $d\eta=0$. By Poincaré lemma, there is a linear $k$-form $\omega'$ such that $d\omega'=\eta$. It follows that $\int_{K}(\omega-\omega')=0$ for any closed $k$-dimensional submanifold $K$. Repeating the same argument, we get that $\omega-\omega'$ is linear; whence the result follows.

Let me just record a few notes on Anton's very nice answer since my comments are already overflowing. This is mostly for my own benefit, and I'm making it community wiki.

Last things first, the argument proceeds by first showing that the "exact density" $\eta$ is linear, and then bootstraps to show that $\omega$ is linear too. In showing that $\eta$ is linear, the only fact used (besides the fact that $\eta$ is a density) is that $\eta$ integrates to 0 over any closed submanifold by the usual Stokes' Theorem-type argument: $\int_S \eta = \int_{\partial S} \omega = \int_{\emptyset} \omega = 0$. This implication is used twice in the bootstrap argument, applied first to $\eta$ and then to $\omega - \omega'$: very elegant.

To show that $\eta$ is linear, Anton formulates a very nice criterion of linearity, that $\sum_{i=1}^n s_i = 0$ should imply that $\sum_{i=1}^n \eta(s_i) = 0$. I believe that we can take $n=3$ here (in particular, let's forget that I used $n$ to denote the dimension of the ambient space way back in the problem statement).

The heart of the proof is to apply the Burago-Ivanov theorem. The point is that the integral in the theorem statement becomes just the sum $s_1 + \dots + s_n = 0$. So the theorem gives us a closed submanifold $S$ whose tangents are concentrated along the $s_i$, with appropriate weighting, except for a set of Euclidean measure $< \epsilon$. Following Anton's comment (with updated notation),

$\int_S \eta = \int_{S_1} \eta + \int_{S_2} \eta = \sum \eta(s_i) + O_{\eta}(\epsilon)$

where $S_1 \subset S$ is the set of points where $S$ is tangent to some $s_i$ and $S_2 \subset S$ is the remainder. Now, as advertized, we use the fact that $\eta$ integrates to 0 over the closed submanifold $S$: $\int_S \eta = 0$, so $\sum \eta(s_i) = O_{\eta}(\epsilon)$. Since $\epsilon>0$ was arbitrary, we have $ \sum \eta(s_i) = 0$.

The Burago-Ivanov theorem was a little intimidating for me. There is a measure on the Grassmanian and a weird $\Lambda^k(\mathbb{R}^n)$ - valued integral over the Grassmanian (note that "$x$" in the integral denotes the identity function when the Grassmanian is regarded as a subspace of $\Lambda^k(\mathbb{R}^n)$, or alternatively it denotes the inclusion map $G(k,n) \to \Lambda^k(\mathbb{R}^n)$). But it's not so bad. The point of the theorem is that it's actually easy to construct measures on the Grassmanian. The "geometric" way is to take some surface $S$ and pushforward its Euclidean surface area along the Gauss map: the resulting measure tells you how much of the surface is parallel to a given plane represented by a point in the Grassmanian. Burago and Ivanov call this the "weighted Gaussian image" of the surface $S$. The "algebraic" way is to simply pick a finite number of unit $k$-vectors $u_1, \dots, u_N$, and assign weights $\lambda_i$ to them to form a discrete measure $\mu$ with $\mu(u_i) = \lambda_i$. For example, Anton constructs a measure this way in his argument.

We can ask whether an algebraically-constructed measure also arises from the geometric construction, as a weighted Gaussian image. Any finitely-supported measure arises geometrically by taking a bunch of disconnected shapes of appropriate areas which lie in the appropriate planes. But according to Burago and Ivanov, we can do better than this (if we're willing to allow an $\epsilon$ of room): we can can construct such a surface whose boundary is controlled by the integral of the measure (which, again, for a discrete measure is a weighted sum of its supporting $k$-vectors). In particular, if the integral / sum comes out to zero, we can take the surface to be closed: this correspondence between an algebraic condition (related to linearity) and a geometric condition amenable to the Stokes' criterion is exactly what we need.

The $\epsilon$ of room corresponds to parts of the surface whose tangent planes weren't supported by our measure. This is needed because for a general set of planes we can't quite construct a closed polytope with sides parallel to exactly those planes. For example, the planes might be linearly independent, so there's no lower-dimensional subspace for them to intersect along. Given that there are such geometric obstructions, it's pretty remarkable that the Burago-Ivanov theorem actually holds! What does such surfaces look like? Burago and Ivanov provide one example: consider a triangular prism: by stretching the prism on one axis and shrinking on the other, you can shrink the end caps to be arbitrarily small while maintaining constant area of the other sides. This is the sort of thing that has to happen.