Yes, I think you can make an example like this (for $p=2$, but it generalizes).
Let $R$ be the group ring $\mathbb Z[C_2]=\mathbb Z[x]/(x^2-1)$. Make a chain complex of free $R$-modules $$ M_0 \leftarrow M_1 \leftarrow M_2 $$ as follows.
$M_0=R$.
$M_1$ has a basis $(a_n)$ indexed by $n\ge 0$.
$M_2$ has a basis $(b_n)$ also indexed by $n\ge 0$.
$\partial a_0=x-1$, $\partial a_n=0$ when $n>0$, $\partial b_n=(1+x)a_n+(1-x)a_{n+1}$.
The homology is such that $H_0\cong\mathbb Z$ (trivial action), $H_2=0$, and $H_1$ has exponent $4$.
Now build a cell complex with free $C_2$-action, having this as its complex of cellular chains. There is one orbit of $0$-cells, say $e^0$ and $xe^0$. There are $1$-cells $e^1_n$ and $xe^1_n$. The cell $e^1_0$ is attached to $e^0$ and $xe^0$, while for $n>0$ both ends of $e^1_n$ are attached to $e^0$. There are $2$-cells $e^2_n$ and $xe^2_n$, with the attaching map for $e^2_n$ representing the appropriate $1$-dimensional homology class of this $1$-skeleton.
