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.

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. $X$ has a $2$-sheeted covering space $\tilde X$, and if $H_i(\tilde X)$ is a $\mathbb Z[1/2]$-module then it splits as a direct sum of the $+1$ and $-1$ ``eigenspaces'' for the action of the covering transformation, which are then $H_i(X)$ and $H_i(X;\mathcal L)$. These two parts need have nothing to do with each other.