There are certain cases where the answer is yes.

If $Y$ is a free $G$-complex, then the Cartan-Leray spectral sequence of the regular cover $Y\to Y/G$ is of the form
$$ H^p(BG;H^q(Y;A))\Rightarrow H^{p+q}(Y/G;A).$$
For instance, in the extreme case that $Y$ has the $A$-cohomology of a point, one has $H^\ast(BG;A)\cong H^\ast(Y/G;A)$. 

There should be other cases where it is possible to draw conclusions about vanishing of group cohomlogy by working backwards through this spectral sequence.

Reference: Ken Brown, *Cohomology of groups,* Section VII.7.

**Added:** In particular, if $Y$ is a $\mathbb{Z}$-acyclic complex with a free $G$-action, then the cohomological dimension of $G$ is less than or equal to $\mathrm{dim}(Y)$. See the MR of 

Howie, James, *Bestvina-Brady groups and the plus construction.* 
Math. Proc. Cambridge Philos. Soc. 127 (1999), no. 3, 487–493

for an example application.