Let $A$ be a subgroup of a group $G$. Then since $A$ is a subgroup of the fundamental group $\pi_1(K(G,1))=G$, there is a covering space $p\colon Y\to K(G,1)$ with $p_*(\pi_1(Y))=A$. So the homology of $Y$ should be completely determined by $A$ and $G$. Suppose that $A$ and $G$ is known, how can one compute $H_*(Y)$, the homology groups of $Y$?
[Math] Homology of Covering Spaces
at.algebraic-topologyfundamental-grouphomology
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For me it is easier to work with cohomology (just for psychological reasons). Also, I will distinguish the representation $\rho$ from the local system $V$ with fibres ${\mathbb C}^2$ that it gives rise to. So where you would write $H^1(F,\rho)$ I will write $H^1(F,V)$. I will let $\overline{V}$ denote the complex conjugate local system to $V$. (So it is the same underlying local system of abelian groups, but we give it the conjugate action of $\mathbb C$.)
The Hermitian pairing on the fibres of $V$ and $\overline{V}$ gives a pairing of local systems $V \times \overline{V} \to \mathbb R$, where $\mathbb R$ is the constant local system with fibre the real numbers. If you like we can think of this as an $\mathbb R$-linear map $V\otimes_{\mathbb C}\overline{V} \to \mathbb R.$ This pairing will induce a map on cohomology $H^2(F,V\otimes_{\mathbb C}\overline{V}) \to H^2(F,\mathbb R)$.
There will also be a cup product $H^1(F,V) \times H^1(F,\overline{V}) \to H^2(F, V\otimes_{\mathbb C} \overline{V})$. Composing this with the previous map on $H^2$ gives your twisted cup product $H^1(F,V)\times H^1(F,\overline{V}) \to H^2(F,\mathbb R)$.
This gives one perspective on your construction. To compute it, write down the twisted cochains $C^{\bullet}(\tilde{F})\otimes_{\mathbb Z[\pi_1(F)]}\mathbb C^2$, then write down the cup-product $$(C^{\bullet}(\widetilde{F})\otimes_{\mathbb Z[\pi_1(F)]}\mathbb C^2 ) \times (C^{\bullet}(\widetilde{F})\otimes_{\mathbb Z[\pi_1(F)]}\mathbb C^2) \to C^{\bullet}(\widetilde{F})\otimes_{\mathbb Z[\pi_1(F)]} \mathbb R^2 = C^{\bullet}(F,\mathbb R).$$ The cup product will just be given by the usual formula, and then you will also pair the $\mathbb C^2$ parts of the cochains using the hermitian pairing.
Hopefully you can follow your nose and do this explicitly for the torus. Then you can just dualize everything to get to the homology version.
This is certainly false! One should look at the corresponding group theory problem: any group $G$ is the fundamental group of a manifold $M$, and the first homology of $M$ is $\pi_1(M)^{ab} = G^{ab}$. Thus the problem becomes: given a group $G$ and a finite index subgroup $H$ of index $p$, does $G^{ab}$ torsion free imply that $H^{ab}$ is torsion free ( edit $p$-torsion free)? (It is unclear whether you are insisting that the $p$-covers be Galois, which would correspond to insisting that $H$ is normal, but both variations have negative answers.)
For an explicit example, let $G = \langle x,y \ | \ [x,y]^2 \rangle$. Then $G^{ab} = \mathbf{Z}^2$, but the homomorphism $G \rightarrow \mathbf{Z}/2$ sending $x$ and $y$ to $1$ has kernel $H = \langle a,b,c \ | \ (cb^{-1}a^{-1})^2, (c^{-1}ba)^2 \rangle$, where $a = yx^{-1}$, $b = x^2$, and $c = xy$. In particular, we see that $H^{ab} = \mathbf{Z}^2 \oplus \mathbf{Z}/2$.
A previous version discussed knot complements and the Alexander polynomial, but I had missed read "torsion free" for "$p$-torsion free", and so the example does not apply.
One positive remark in the direction of your question: If you are assuming that $H$ is normal, then $H^{ab}$ is $p$-torsion free if the rank of $G^{ab}$ is less than two. This is because the latter condition implies that the $p$-completion of $G$ is cyclic, a condition which is inherited by a normal subgroup of index $p$.
The methods used to compute homology vary considerably depending on what information you have. If you have a presentation for the group, you can use the Reidemeister--Schreier algorithm to compute a presentation of $H$, from which it is easy to compute $H^{ab}$. The more you understand the geometry of the situation, however, the better.
Best Answer
This is not a trivial problem. A favorite example of mine is the case of a knot complement, $S^3\setminus K$. (It is known that these are Eilenberg-Maclane spaces.) If you pick $A$ to be the commutator subgroup of $\pi_1(S^3\setminus K)=G$, then $Y$ is called the universal Abelian cover, and $H_1(Y;\mathbb Q)$ turns out to be a torsion $\mathbb Q[t,t^{-1}]$-module, called the Alexander module of the knot. The order of the Alexander module is called the Alexander polynomial. One can calculate the Alexander module from the fundamental group using Fox calculus.