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The answer is NO, in general. To take 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. Any abelian quotient of this congruence closure may easily be seen to consist only of two and three torsion, and hence the kernel of $\Gamma$ to ${\mathbb Z}/5{\mathbb Z}$ cannot be a congruence quotient.