To rephrase Alex Bartel's answer geometrically:
the quotient of the hyperbolic plane by $\Gamma(2)$ is a 3-punctured sphere, whose homology
is free abelian of rank 2.

**Added comment**: To clarify the relationship of the groups: neither $SL(2,\mathbb Z)$ nor $\Gamma(2)$ act effectively on $\mathbb H^2$: the center of $SL(2,\mathbb Z)$, generated by $-I$, acts trivially. So, $\pi_1(\mathbb H^2)/\Gamma(2)$ is indeed the commutator subgoup of $SL(2,\mathbb Z)$, but as Alex Bartel says, it has index 12 in $SL(2,\mathbb Z)$. $\Gamma(2)$ itself (the subgroup of $SL(2,\mathbb Z)$ congruent to the identity mod 2) is isomorphic to the product of this fundamental group with the order 2 group
 $\left < -I \right > $.

 Any space has a universal abelian cover, whose group of deck transformations is its homology; the homology of the universal abelian cover is what you're asking for. The universal abelian cover is easier to see for
the punctured torus, which has the same fundamental group --- it's just the plane minus holes 
for the punctures. The homology of this cover is coordinatized by linking number with the holes.

To picture the universal abelian cover of the 3-punctured sphere: enlarge the punctures
to make boundary components, forming a pair of pants.  Cut the pair of pants along 3 seams into two hexagons.
Use a template consisting of the plane divided into equilateral triangles with
an alternating coloring, say red and blue; color the hexagons red and blue.
Above each red triangle, put a red hexagon above it in $\mathbb R^3$ so that three of the hexagon's edges
project to the midpoints of edges of the red triangles, and the other three
connect the top of one vertical edge to the bottom of the counterclockwise vertical edge.
Arrange the blue hexagons in the same way, over the blue triangles.
Half of the edges of hexagons string together to form lines that weave in and out, in
a hexagonal weave: these cover the three boundary components of the pair of pants.