Wikipedia tells us that most cubic graphs have a Hamilton cylce (for instance the proportion of Hamiltonian graphs among the cubic graphs on $2n$ vertices converges to 1 as $n$ goes to infinity) but is also kind enough to provide us with some pictures of non-Hamiltonian cubic graphs:
Staring at these graphs it is easy to see a pattern: in all three cases there is a large cycle $C$ such that all the points not on $C$ have all three of their neighbours on $C$ instead of being connected to eachother. (In the bottom two graphs this is most obvious.) Since the same is somewhat vacuously true for cubic graphs that do have a Hamilton cycle my question is:
Is it true that in every connected bridgeless cubic graph there is a cycle $C$ such that each vertex is either on $C$ or has all three of its neighbors on $C$?
(Of course also many vertices on $C$ have all their neighbors on $C$ so in spite of writing 'either' I just mean the ordinary, inclusive or here, not the exclusive or.)
Edit: the graphs in the pictures are called 'Tutte graph', 'Coxeter graph' and 'Petersen graph' respectively.
Even though this question seems to be of no interest to anyone I still thought it would be good form to post here a counterexample I found since posting:
Suppose (aiming for a contradiction) that the graph contains a cycle $C$ of the form described in the question. Picking any of the red edges it is easy to conclude that it is absolutely necessary that this edge belongs to $C$. Now by symmetry the same arguments apply to the other red edges and hence all 8 of them must be part of $C$. This in turn implies that the green vertices are not on $C$ but have exactly 2 of their neighbors on $C$ instead of 3, contradicting the presumed nature of $C$.