The answer is NO, in general. For a specific counterexample, let $\Gamma$ be the principal congruence subgroup of level two in $SL(2,{\mathbb Z})$. Then, $\Gamma$ modulo $\pm 1$, is the free group on two generators, and hence there is a homomorphism from $\Gamma$ onto ${\mathbb Z}/5{\mathbb Z}$ (the latter realised as the quotient of the free abelian group on two generators modulo a suitable subgroup). The congruence closure of $\Gamma$ may easily be shown to be the product $$SL_2(2{\mathbb Z}_2)\times SL(2,{\mathbb Z}_3) \times \prod _{p\neq 2,3} SL(2,{\mathbb Z}_p)$$where ${\mathbb Z}_p$ denotes the ring of $p$-adic integers. Using the fact that for $p\geq 5$, the group $SL_2({\mathbb Z}_p)$ is its own commutator, it is easy to see that any Abelian quotient of this congruence closure may easily be seen to consist only of two and three torsion. Hence the kernel of $\Gamma$ to ${\mathbb Z}/5{\mathbb Z}$ cannot be a congruence subgroup.
[Edit] I see that the link provided by Matthias Wendt answers this question completely.