Is there a non-bipartite hamiltonian cubic graph on $n$ vertices with no $(n-1)$-cycle?

Question

Is there a cubic (3-regular) graph $G$ on $n$ vertices such that:

• $G$ is hamiltonian
• $G$ has no $(n-1)$-cycles
• $G$ is not bipartite

My computer tells me that there are none on up to $24$ vertices.