I think in general the answer is no. Take the map $\mathbb P^2\to \mathbb P^2$ given by $[x_0,x_1,x_2]\mapsto [x_0^2, x_1^2, x_2^2]$. This is the quotient map for the action of $G=\mathbb Z_2^2$ given by $[x_0,x_1,x_2]\mapsto [x_0,\pm x_1, \pm x_2]$. Since $H^2(\mathbb P^2, \mathbb Z)=\mathbb Zh$, where $h$ is the class of a line, $G$ acts trivially on $H^2(\mathbb P^2, \mathbb Z)$. On the other hand, the pullback of $H^2(\mathbb P^2, \mathbb Z)$ is the subgroup generated by $2h$.
ADDED: this example shows that the answer is no whenever $h^2(X)=h^2(X/G)$, because the intersection form on $H^2(X,\mathbb Z)$ is unimodular, while the intersection number of two classes coming form $X/G$ is divisible by $|G|$. (At least when $X/G$ is a smooth surface, I'm not sure what happens when $X/G$ is singular.)
I think in general the answer is no. Take the map $\mathbb P^2\to \mathbb P^2$ given by $[x_0,x_1,x_2]\mapsto [x_0^2, x_1^2, x_2^2]$. This is the quotient map for the action of $G=\mathbb Z_2^2$ given by $[x_0,x_1,x_2]\mapsto [x_0,\pm x_1, \pm x_2]$. Since $H^2(\mathbb P^2, \mathbb Z)=\mathbb Zh$, where $h$ is the class of a line, $G$ acts trivially on $H^2(\mathbb P^2, \mathbb Z)$. On the other hand, the pullback of $H^2(\mathbb P^2, \mathbb Z)$ is the subgroup generated by $2h$.