In principle one can get an exact answer to this question by a lot of case checking and a quantitative version of the following result.

**Theorem**. Let $G$ be a finite group such that $r_2(g) \geq \varepsilon |G|$ for some $g \in G$ and $\varepsilon > 0$. Then $G$ contains a subgroup $H$ of index $O_\varepsilon(1)$ which is $2$-step nilpotent with $|[H,H]| \ll_\varepsilon 1$.

Roughly speaking, this says that a group $G$ with a large value of $r_2$ must be ``bounded-by-abelian-by-bounded'', which should be enough structure to then do a case analysis, at least in principle.

*Proof*. Since all conjugates of $g$ have a value of $r_2$ greater than or equal to $\varepsilon |G|$, there are at most $1/\varepsilon$ conjugates of $g$. Thus the centraliser $C_G(g)$ of $g$ has index $O_\varepsilon(1)$ (cf., Derek Holt's comment). By the pigeonhole principle, there is a coset $x C_G(g)$ of $C_G(g)$ which contains $\gg_\varepsilon |G|$ solutions $xt, t \in C_G(g)$ to the equation $(xt)^2 = g$; by shifting $x$ without loss of generality we may also assume that $x^2=g$. From $x^2 = g$ we have $x = x^{-1} g$ and hence $(xt)^2 = xtx^{-1} gt = xtx^{-1} tg$ since $t \in C_G(g)$. Thus the equation $(xt)^2=g$ is equivalent to
$$xtx^{-1} = t^{-1}.\quad (1)$$
So (1) holds for $\gg_\varepsilon |G|$ choices of $t \in G$. In particular, for $\gg_\varepsilon |G|^2$ pairs $(t,s) \in G^2$, we have
$$ xtx^{-1} = t^{-1}; \quad xts x^{-1} = s^{-1} t^{-1}$$
which implies
$$ xsx^{-1} = t s^{-1} t^{-1}.$$
By Cauchy-Schwarz, this implies that there are $\gg_\varepsilon |G|^3$ triples $(s,t_1, t_2)$ such that
$$ t_1 s^{-1} t_1^{-1} = t_2 s^{-1} t_2^{-1}$$
and so there are $\gg_\varepsilon |G|^2$ solutions to the equation $xyx^{-1} = y$. In other words, the commuting probability of $|G|$ is $\gg_\varepsilon 1$. The claim then follows from a theorem of Neumann (see Theorem 2.4 of this paper of Eberhard). $\Box$

I suspect that in the case $\varepsilon > 1/2$ one can work a little harder and show that one can assume without loss of generality that the abelianisation $H/[H,H]$ is $2$-torsion, so that $G$ is ``virtually'' a $2$-torsion group in some weak sense. On the other hand, once $\varepsilon$ reaches $1/2$ one can have non-$2$-torsion behavior. For instance, let $H$ be an arbitrary abelian group, let $g$ be an arbitrary element of $H$ [EDIT: as pointed out by Will below, one needs to assume that $g$ is of order $2$], and consider the group $\langle H, e \rangle$ where $e$ is subject to the relations $e^2 = g$, $ehe^{-1} = h^{-1}$ for all $h \in H$ (this is a sort of twisted dihedral group, I don't know the official name for it). Then $G = \langle H,e \rangle$ is a group containing $H$ as an index $2$ subgroup with $r_2(g) = |G|/2$.