Let $H$ be a group, if its abelianization is $\mathbb{Z}_2$ does this mean that $H$ has torsion?

No. Take $G$ to be an infinite, torsion-free perfect group. Let $x$ be a nontrivial element of $G$. Take the amalgam of $G$ with $\mathbf{Z}$ by identifying $x$ with $2$: $$H=\langle G,c\mid c^2=x\rangle.$$ As an amalgam of two torsion-free groups, it is torsion-free. From the presentation, the abelianization is clearly of order 2, generated by the image of $c$.

There are many choices for $G$, including finitely presented ones, for instance $\langle s,t\mid (st)^2=s^3=t^7\rangle$.

I believe the answer should be “no”. What I have is not an example, but I suspect that something along these lines could be massaged into an example.

The group $E\Pi\operatorname{Aut}(F_3)$, is torsion-free (a theorem of Glover and Jensen) and given by the following presentation. Throughout, $i,j,k$ will be pairwise distinct elements of $\{1,2,3\}$.

$$E\Pi\operatorname{Aut}(F_3) = \langle \rho_{ij} \mid [\rho_{ij},\rho_{ik}], \rho_{ik}\rho_{jk}\rho_{ij} = \rho_{ij}\rho_{jk}\rho_{ik}^{-1}\rangle.$$

Notice that the latter relator abelianizes to saying the image of $\rho_{ik}$ is equal to the image of $\rho_{ik}^{-1}$, so the abelianization of $E\Pi\operatorname{Aut}(F_3)$ is $C_2^6$.