Take $k$ long paths between two vertices $a,b$. This graph is already 2-connected. Now add some edges to make it $k$-regular so that every edge joins only interior vertices of the same path (it is not hard). Then $G\setminus\{a,b\}$ has more connected components than any graph with Hamiltonian cycle with two vertices removed.

Take $k>3$ long paths between two vertices $a,b$. This graph is already 2-connected. Now add some edges to make it $k$-regular so that every edge joins only interior vertices of the same path (it is not hard). Then $G\setminus\{a,b\}$ has more connected components than any graph having Hamiltonian path with two vertices removed.