No. Fix $p\ge 2$. Take the group $$G=\{M(x,y,z;n):(x,y,z)\in\mathbf{Z}[1/p],n\in\mathbf{Z}\}$$where $$M(x,y,z;n)=\begin{pmatrix}1 & x & z \\ 0 & p^n & y\\ 0 & 0 & 1\end{pmatrix}$$
and $K$ the set of such $M(x,y,z;n)$ for $n=0$, and identify $\mathbf{Z}$ to powers of $M(0,0,0,1)$.
Then $G$ is finitely generated (namely by $\{M(0,0,0;1),M(1,0,0;0),M(0,1,0;0)\}$), $K$ is not finitely generated, and indeed the centralizer of $M(0,0,0;1)$ in $K$ is not finitely generated (isomorphic to the abelian group $\mathbf{Z}[1/p]$).