Killing cohomology of surfaces in finite covers Let $S$ be a closed orientable surface and $R$ a commutative ring.  Given a nonzero element $\alpha \in H^1(S;R)$ is there a finite cover $p : \tilde{S} \to S$ such that $p^*(\alpha) = 0 \in H^1(\tilde{S}; R)$?  

No. This has nothing to do with $S$ being a surface. In what follows all we assume is that $S$ can run the classification theory of covers (path-connected, locally path-connected, semilocally simply connected). It holds for any CW complex.
There is a natural isomorphism $H^1(S;G) \cong \text{Hom}(\pi_1 S, G)$ for any $S, G$. Given any element $\phi \in \text{Hom}(\pi_1 S, G)$, if $f: S' \to S$ is a covering space, the correspinding map $\phi f_* = 0$ if and only if $f$ factors through the cover corresponding to $\text{ker}(\phi) \subset \pi_1 S$. (We use here the classification of covering spaces over reasonable spaces.)
If $\text{ker}(\phi)$ has infinite index, then the cover corresponding to $\text{ker}(\phi)$ is an infinite cover. For instance, if $G$ is infinite and $\phi$ is surjective (or at least its image has finite index), then you cannot kill $\phi$ in a finite cover. 
Because every subgroup of $\Bbb Z$ is zero or infinite, in particular, you cannot kill any nontrivial class of $H^1(S;\Bbb Z)$ in a finite cover.
