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.
The answer is no by Yves' comments. Let me add that there are plenty of explicit constructions of closed hyperbolic 3--manifolds with finite homology, and this is a generic phenomenon (for example random Heegard gluings have zero first Betti number and are hyperbolic and numerical experiments on the census manifolds exhibit an overwhelming proportion of manifolds with zero first Betti number). This hints to there being no hope to get a classification.
For various results about hyperbolic rational homology spheres (probabilistic, numerical, explicit constructions of infinite families) see for example the papers of Nathan Dunfield and coauthors:
Best Answer
I don't think that
Certainly, torsion in Cech cohomology has been ruled out for a compact subset. The "usual" universal coefficient formula, relating Cech cohomology to $\operatorname{Hom}$ and $\operatorname{Ext}$ of Steenrod homology, is not valid for arbitrary compact subsets of $\Bbb R^3$ (although it is valid for ANRs, possibly non-compact). The "reversed" universal coefficient formula, relating Steenrod homology to $\operatorname{Hom}$ and $\operatorname{Ext}$ of Cech cohomology is valid for compact metric spaces, but it does not help, because $\operatorname{Ext}(\Bbb Z[\frac1p],\Bbb Z)\simeq\Bbb Z_p/\Bbb Z\supset\Bbb Z_{(p)}/\Bbb Z$, which contains $q$-torsion for all primes $q\ne p$. (Here $\Bbb Z_{(p)}$ denotes the localization at the prime $p$, and $\Bbb Z_p$ denotes the $p$-adic integers. The two UCFs can be found in Bredon's Sheaf Theory, 2nd edition, equation (9) on p.292 in Section V.3 and Theorem V.12.8.)
The remark on $\operatorname{Ext}$ can be made into an actual example. The $p$-adic solenoid $\Sigma$ is a subset of $\Bbb R^3$. The zeroth Steenrod homology $H_0(\Sigma)$ is isomorphic by the Alexander duality to $H^2(\Bbb R^3\setminus\Sigma)$. This is a cohomology group of an open $3$-manifold contained in $\Bbb R^3$, yet it is isomorphic to $\Bbb Z\oplus(\Bbb Z_p/\Bbb Z)$ (using the UCF, or the Milnor short exact sequence with $\lim^1$), which contains torsion. Of course, every cocycle representing torsion is "vanishing", i.e. its restriction to each compact submanifold is null-cohomologous within that submanifold.
By similar arguments, $H_i(X)$ (Steenrod homology) contains no torsion for $i>0$ for every compact subset $X$ of $\Bbb R^3$.
It is obvious that "Cech homology" contains no torsion (even for a noncompact subset $X$ of $\Bbb R^3$), because it is the inverse limit of the homology groups of polyhedral neighborhoods of $X$ in $\Bbb R^3$. But I don't think this is to be taken seriously, because "Cech homology" is not a homology theory (it does not satisfy the exact sequence of pair). The homology theory corresponding to Cech cohomology is Steenrod homology (which consists of "Cech homology" plus a $\lim^1$-correction term). Some references for Steenrod homology are Steenrod's original paper in Ann. Math. (1940), Milnor's 1961 preprint (published in http://www.maths.ed.ac.uk/~aar/books/novikov1.pdf), Massey's book Homology and Cohomology Theory. An Approach Based on Alexander-Spanier Cochains, Bredon's book Sheaf Theory (as long as the sheaf is constant and has finitely generated stalks) and the paper
As for torsion in singular $4$-homology of the Barratt-Milnor example, this is really a question about framed surface links in $S^4$ (see the proof of theorem 1.1 in the linked paper).