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Gunnar Brinkmann informs me that this paper constructs planar triangulations where every hamiltonian cycle misses all the edges of many triangles. Some of the examples are even 5-connected.

Carol Zamfirescu and Gunnar Brinkmann have informed me that this paper answers the question.

This is a NEW EDITION using the condition that two edges incident with a vertex not in the dominating circuit cannot be attached to consecutive vertices in the dominating circuit. I tried 200 million …

Yes, every connected cubic graph is 3-almost-Hamiltonian. Replace each edge by two parallel edges then follow an Eulerian circuit. In the case of a bridgeless cubic graph, you can add a perfect match …

Take any $k\ge 1$ and four disjoint sets of vertices $A,B,C,D$ with $|A|=|D|=k+2$, $|B|=|C|=k$. Completely join $A$ to $B$, $B$ to $C$ and $C$ to $D$. This gives a bipartite graph of minimum degree $ …

If a full set of $n$ edges is inserted between the cycles, what you have is a "cycle permutation graph". According to this paper, there are nonhamiltonian cycle permutation graphs for all odd $n\ge 9$ …

Tony's method works in the general case. Let $e_1, e_2, \ldots, e_N$ be the edges that are added to $G$, in the order they are added, to make the complete graph. Choose an arbitrary hamiltonian cycl …

Bill Kocay found a more direct combinatorial method to reconstruct the number of hamiltonian cycles and some other spanning subgraphs. It is in his paper "Some new methods in reconstruction theory", …