The answer is No in general. Let $n\geq 3$ be odd (it is not necessary that $n$ be odd) and suppose $G=\mathrm{SL}_n({\mathbb Z})$. There exists a subgroup $\Gamma \subset \mathrm{SL}_n({\mathbb Z})$ 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, $\mathrm{SL}_n({\mathbb Z})$ has the congruence subgroup property which means that we have the following inclusion of the profinite completions:
$$\widehat {\mathrm{SL}_n({\mathbb Z})}= \prod _{q \; \mathrm{prime}} \mathrm{SL}_n({\mathbb Z}_q)\supset \widehat {\Gamma} \supset \prod _{p\in S} U_p \times \prod _{ \ell \notin S} \mathrm{SL}_n({\mathbb Z}_\ell),$$
where $S$ is a finite set of primes, $U_p$ is an open subgroup of finite index in $\mathrm{SL}_n({\mathbb Z}_p)$, and $\ell$ runs through primes in the complement of $S$. Since for infinitely many $\ell$ (for example, all $\ell$ with $\ell\equiv 1 \; (\mathrm{mod}\;n)$), the group $\mathrm{SL}_n({\mathbb Z}_\ell)$ has $n$-th roots of unity in the centre, it follows that the profinite completion of $\Gamma $ is not centreless.
