Finite groups with lots of conjugacy classes, but only small abelian normal subgroups?

No: there exists a sequence of finite groups with commuting probability bounded away from 0 but with no abelian (normal) subgroup of bounded index.

Fix a prime power $q$. Consider the "higher Heisenberg" group $G_n$ of order $q^{2n+1}$ consisting of those square matrices of size $n+2$ over $\mathbf{F}_q$ of the form $$P(u,v,x)=\begin{pmatrix}1 & u & x\\ 0 & I_n & v\\ 0 & 0 & 1\end{pmatrix},$$ where $u$ is a row, $v$ is a column, and $x$ a scalar. If we endow $\mathbf{F}_q^{n^2}$ with the symplectic product $\langle u\oplus v,u'\oplus v'\rangle=uv'-u'v$, then we see that the centralizer of $P(u,v,x)$ is the set of $P(u',v',x')$ such that $\langle u\oplus v,u'\oplus v'\rangle=0$. In particular, this is a subgroup of index $q$ (unless $(u,v)=(0,0)$ in which case $P(u,v,x)$ is central). So the probability that two elements commute is $\ge 1/q$ [actually it's $1/q+q^{-2n}(1-1/q)$, if I'm correct].

On the other hand, the largest cardinal of an abelian subgroup in $G_n$ is $q^{n+1}$ (since $n$ is the largest dimension and hence $q^n$ is the largest cardinal of an isotropic subspace in the symplectic space $\mathbf{F}_q^{2n}$). So the minimal index of an abelian subgroup in $G_n$ is $q^n$. Thus, $q$ being fixed ($q=2$ is fine) and $n$ tending to infinity, we get the required example.