Easy infinite subclass of cubic graphs for Hamiltonian cycle problem
Yes, there is an infinite class of 2-connected cubic graphs on which Hamilton Cycle has a polynomial-time algorithm. Further, there is a such a class that contains infinitely many Hamiltonian graphs and infinitely many non-Hamiltonian graphs, which I think is a decent definition of "non-trivial".

First, let $H_n$ be the union of a $2n$-cycle on vertices $\{1, \dots, 2n\}$ and the edges $\{(i, i+n)\mid 1\leq i\leq n\}$. $H_n$&nbsp;is 3-regular, 2-connected and has an obvious Hamiltonian cycle.

Now, let $G$ be a 2-connected cubic graph that has no Hamiltonian cycle. Such a graph must exist, since the Hamiltonian Cycle problem is NP-complete on 2-connected cubic graphs, so must have both "yes" and "no" instances. Fix an edge $xx'\in G$. For any graph&nbsp;$H_n$ pick any edge $yy'\in H_n$ and let $H'_n$ be the graph made by taking $G\cup H_n$, deleting the edges $xx'$ and $yy'$ and adding edges $xy$ and $x'y'\!$.

$H'_n$ is 3-regular and 2-connected but I claim that it has no Hamiltonian cycle. Any Hamiltonian cycle $C$ in $H'_n$ must enter the copy of&nbsp;$G-xx'$ at&nbsp;$x$ and leave at&nbsp;$x'$ (or vice-versa). But then $C$ must contain a Hamiltonian path&nbsp;$P$ of $G-xx'$ that begins at&nbsp;$x$ and ends at&nbsp;$x'\!$. However, no such Hamiltonian path can exist, since $P\cup\{xx'\}$ would be a Hamiltonian cycle of&nbsp;$G$, but $G$ was chosen to have no Hamiltonian cycles.

So the desired class of graphs is $\mathcal{H} = \{H_n\mid n&gt;1\}\cup\{H'_n\mid n&gt;1\}$. It remains to show that the Hamiltonian Cycle problem can be solved in polynomial time for graphs in&nbsp;$\mathcal{H}$. Observe that, for every&nbsp;$n$, every edge of $H_n$ is on a 4-cycle (there are 4-cycles of the form $i$, $i+1$, $i+n+1$, $i+n$, $i$). However, the edges $xy$ and $x'y'$ are not on any 4-cycle of&nbsp;$H'_n$. We can test that every edge of a graph is in a 4-cycle in time $\mathcal{O}(n^4)$.



I edited to changed the construction slightly, with two benefits. First, the algorithm runs in time $\mathcal{O}(n^4)$ instead of $\mathcal{O}(n^{|V(G)|})$. Second, there's an actual correctness proof; the old algorithm relied on the assumption that $G-xx'$ is not a subgraph of any $H_n$, which was probably true but really needed to be proven.