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The answer is No in general. Let $n\geq 3$ be odd (it is not necessary that $n$ be odd) and suppose $G=SL_n({\mathbb Z})$. There exists a subgroup $\Gamma \subset SL_n({\mathbb Z})$ of finite index which is torsion free and centreless (the centre can only be $\pm 1$ and because $n$ is odd the centre can only be trivial). However, $SL_n({\mathbb Z})$ has the congruence subgroup property which means that the profinite completion of $\Gamma$ contains a group of the form $\prod _{p\in S} U_p \times \prod _{ l \notin S} SL_n({\mathbb Z}_l)$, where $S$ is a finite set of primes, $U_p$ is an open subgroup of finite index in $SL_n({\mathbb Z}_p)$, and $l$ runts through primes in the complement of $S$. Since for infinitely many $l$, $SL_n({\mathbb Z}_l)$ has $n$-th roots of unity in the centre, it follows that the profinite completion of $\Gamma $ is not centreless.