Does there exist a surjective homomorphism from $(\mathbb R,+)$ to $(\mathbb Q,+)$ ?

Sure.

Fix a Hamel basis $B$ for $\Bbb R$ over $\Bbb Q$, fix some $r\in B$ and map $B\setminus\{r\}$ to $0$, and $r$ to $1$. Then you're done.

If you want to avoid the axiom of choice, you can't. It is consistent that every homomorphism from $\Bbb R$ to $\Bbb Q$ is continuous, and therefore its image is connected. But this means that every homomorphism is $0$.

(This is a consequence of "Every set of reals is Lebesgue measurable" as well "Every set of reals has the Baire property", at least in the presence of Dependent Choice. Both of these have been shown consistent with Dependent Choice by Solovay starting with an inaccessible, and later Shelah proved that for the Baire property you do not need an inaccessible cardinal.)

Hint Write $\mathbb R$ as a vector space over $\mathbb Q$. Pick a basis $B$ and define a vector space homomorphism by sending $B$ to $\{ 1\}$.

Show it has the desired property.

Second solution Use Zorn's Lemma to define a maximal homomorphism from a subgroup of $\mathbb R$ to $\mathbb Q$ which is identity on $\mathbb Q$.