This answer complements Oscar Randal-Williams's. First of all, if $G=\mathbb{Z}/p$ acts freely and orientation-preservingly on a closed orientable surface $S$ then $H_{1}(S) =\mathbb{Z}^{2}\oplus (\mathbb{Z}[G])^{n}$. One way to see this is to put the action into a sort of
"normal form". The covering $S\to S/G$ is determined by a classifying map $c:S/G\to BG$, or, equivalently, the corresponding surjective homomorphism $c_{\#}:\pi_{1}(S/G)\to G$. Such a homomorphism can be completely described by listing its values on a standard set of simple closed curves representing a symplectic basis for $H_{1}(S/G)$. One can argue that a system of such curves can be chosen so that all but one curve maps trivially. This result goes back to P.A. Smith, if not further. For details see my old paper
*Allan L. Edmonds*, MR 654478 **Surface symmetry. I**, *Michigan Math. J.* **29** (1982), no. 2, 171--183.
From this normal form, the main claim is now immediate, since the covering is trivial over all except a torus.

On the other hand if $G=(\mathbb{Z}/p)^{2}$ it can happen that the representation has the desired form and also happen that the representation is different. Again one can understand the covering $S\to S/G$ by putting the classifying homomorphism $c_{\#}:\pi_{1}(S/G)\to G$ into standard form. It turns out in this case that up to equivariant homeomorphism and automorphisms of $G$ there are exactly *two* such forms with respect to a suitable system of simple closed curves representing a symplectic basis for $H_{1}(S/G)$. If $G$ has generators $x,y$, then the normal forms are

(1)
$(x,y;1,1;\dots ;1,1)$

(2)
$(x,1; y,1; 1,1;\dots ;1,1)$

Again see my old paper for more details. For Case (1) the corresponding representation on $H_{1}(S)$ again has the form $\mathbb{Z}^{2}\oplus (\mathbb{Z}[G])^{n}$, since the covering $S\to S/G$ is trivial over all but the core torus.

In Case (2), however, things are different. The two cases are distinguished by $c_{*}[S/G]$ in $H_{2}(BG)=\mathbb{Z}/p$. In Case (1) $c_{*}[S/G]\neq 0$. But in Case (2) $c_{*}[S/G]= 0$, as one can visibly see, by surgering curves to create a cobordism from the map $c$ to a map $S^{2}\to BG$, which is necessarily null-homotopic. In this situation we can adapt Oscar's argument to see that $H_{1}(S)$ cannot be of the form $\mathbb{Z}^{2}\oplus (\mathbb{Z}[G])^{n}$.

The spectral sequence of the fibration $S\to S/G \to BG$ leads to a five-term exact sequence of integral homology groups
$$
H_{2}(S/G)\to H_{2}(BG)\to H_{0}(BG; H_{1}(S))\to H_{1}(S/G)\to H_{1}(BG)\to 0.
$$
Because we are in Case (2), the left hand homomorphism is trivial, and the sequence becomes
$$
0\to \mathbb{Z}/p \to H_{0}(BG; H_{1}(S))\to H_{1}(S/G)\to (\mathbb{Z}/p)^{2}\to 0.
$$
Now if $H_{1}(S)=\mathbb{Z}^{2}\oplus (\mathbb{Z}[G])^{n}$ we must have $H_{1}(S/G)=\mathbb{Z}^{n+2}$ by consideration of Euler characteristics. Then the exact sequence becomes
$$
0\to \mathbb{Z}/p \to \mathbb{Z}^{n+2}\to \mathbb{Z}^{n+2}\to (\mathbb{Z}/p)^{2}\to 0,
$$
which is impossible. One could also work with coefficients $\mathbb{F}_{p}$ and reach a similar contradiction. One can describe the $\mathbb{Z}[G]$ module $H_{1}(S)$ more or less explicitly in this case, presumably involving augmentation ideals. Note also that in Case (1) this sequence becomes
$$
0\to \mathbb{Z}^{n+2}\to \mathbb{Z}^{n+2}\to (\mathbb{Z}/p)^{2}\to 0,
$$
which does occur.

3more comments