One approach is to use mod $2$ homology. You know that
$H_i(X;\mathbb Z/2)$ is isomorphic to both $ H_i(X)\otimes \mathbb Z/2\oplus Tor(H_i(X),\mathbb Z/2)$ and $H_i(X;\mathcal L:)\otimes \mathbb Z/2\oplus Tor(H_i(X;\mathcal L),\mathbb Z/2)$. If the integral homology groups are finitely generated, then this gives you what you want.
But if $2$ is invertible in the integral homology then I don't think there's much you can say.