No:  Consider $\mathbb{Z_{2}}$ acting on the universal cover $X=S^{n}$ of $Y=\mathbb{RP}^{n}$ for $n$ even.  As $H^{n}(Y;\mathbb{Z})=\mathbb{Z}_{2}$ and $H^{n}(X;\mathbb{Z})=\mathbb{Z}$, the map $\pi^{*}$ must be zero when restricted to the $n$th cohomology groups.