Inspired by Vor's answer, I want to give a simpler one.
Start with Hamiltonian cycle problem for grid graphs problem which was proven hard by Itai.
It can be easily seen that edge set of a grid graph can be partitioned into 2 disjoint subsets: horizontal and vertical.
So now, we need to weave all horizontal ones into one simple cycle, and weave all vertical ones into another simple cycle.
This is very easy task: for vertical ones, sweep from leftmost to rightmost, just connect any vertical gaps, then connect consecutive x-coordinated vertical line, then connect the lowest leftmost vertex with the highest rightmost vertex. Do similarly for horizontal edges.
Note that the obtained graph is still simple, undirected and satisfies requirement. It is simple because at the last steps of vertical phase and horizontal phase, we deal with two different vertex pairs.
Now, do a similar trick as Vor did. At each vertex, for each of its original incident edge, add $k$ new vertices. As usual, $k$ ahouls be large enough. Lastly, the length of a genuine Hamiltonian cycle should be $2k|V|$. But of course, it is not hamiltonian of the obtained graph.