Let $S$ be a closed oriented surface. Consider the (universal) abelian cover $p \colon S_{ab} \rightarrow S$, i.e. the one whose group of deck transformations is the abelianization of $\pi_1(S,\star)$. My professor told me that $H^1(S_{ab},\mathbb{R}) = 0$. Even more so, he said the abelian cover is the smallest cover for which the first cohomology group vanishes.
EVEN NEWER EDIT:
Here is what I have; I would appreciate any feedback as to correctness of the following arguments.
First of all, $H_1(S,\mathbb{Z}) \cong \mathbb{Z}^{2g}$ ($g$ the genus), so that by the universal coefficients theorem $H_1(S,\mathbb{R}) \cong \mathbb{R}^{2g}$. Thus, the kernel of the Hurewicz map $\pi_1(S) \rightarrow H_1(S,\mathbb{Z})$ is the same as the kernel of the map $\pi_1(S) \rightarrow H_1(S,\mathbb{R})$ and, hence, there is no ambiguity as to which map we use (this refers to one discussion in the comments).
Next, the commutator subgroup of $\pi_1(S)$ is free and for genus $g \geq 2$ is not trivial (let us focus on the higher genus case). Therefore, $$H_1(S_{ab},\mathbb{Z}) \cong \pi_1(S_{ab})^{ab} \cong [\pi_1(S),\pi_1(S)]^{ab}$$ is free-abelian (and not trivial). Let us write this group as $\bigoplus_{j \in J}\mathbb{Z}$, where $J \ne \emptyset$. Then $$H_1(S_{ab},\mathbb{R}) \cong H_1(S_{ab},\mathbb{Z}) \otimes \mathbb{R} \cong \bigoplus_{j \in J} \mathbb{R}.$$ Finally, $$H^1(S_{ab},\mathbb{R}) \cong Hom\big(H_1(S_{ab},\mathbb{Z}),\mathbb{R}\big) \cong Hom\Big(\bigoplus_{j \in J} \mathbb{R},\mathbb{R}\Big)$$ and this group is non-trivial. In conclusion, it seems that the proposed statement in the question is not correct as stated.
EVEN NEWER EDIT THAN THE PREVIOUS EVEN NEWER EDIT:
Sooo, it took me a while to get back on MSE, but here we go again... My professor admitted to accidentally claiming a wrong statement. As verified above, the first cohomology of the abelian cover is often non-trivial. He corrected himself and now claimed that the induced map $p^* \colon H^1(S,\mathbb{R}) \rightarrow H^1(S_{ab},\mathbb{R})$ by pull-back is trivial. Unfortunately, I still don't know how to prove this. I hope my endless amounts of edits have not yet tired you out.