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? (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 counter example of a geometric flavor, let $M$ be a knot complement $S^3 \setminus K$. Then $\pi_1(M)^{ab} = \mathbf{Z}$. However, the torsion subgroup of the covers corresponding to the quotients $\mathbf{Z} \rightarrow \mathbf{Z}/p$ will have torsion subgroups in general (roughly correponding to the "value" of the Alexander polynomial $A_K(t)$ at $t = \zeta_p$ a $p$-th root of unity).

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.

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.