Below are some thoughts when $G$A counterexample is athe planar triangulationdual to the Tutte graph. Most of the credit goes to David Eppstein (see the comments below)
Let $G$ be the dual to the Tutte graph. Then $G$ is a planar triangulation. SupposeTowards a contradiction, suppose that $G$ has a set of edges $C$ such that $G / C$ is a thick-tree $T$ (a tree with multiple-edges). Note that $C$ must contain at least one edge from each face of $G$, since otherwise $T$ will contain a triangle. On the other hand, $C$ cannot contain more than one edge from each face, since otherwise $T$ will contain a loop. Thus, $C$ contains exactly one edge from each face of $G$. ByBy Euler's Theoremformula, this implies that $G / C$ has only two vertices.
Conversely In particular, I had previously claimed that every set $C$ meeting each face once$(G / C)^*$ is a validHamiltonian. Since contraction setand deletion are dual operations, but that is in fact falsewe have $(G / C)^*=G^* \backslash C$. NoteThis implies that suchthe Tutte graph has a set $C$ does exist for every planar triangulation by Peterson's Theorem. See David Eppstein's comments belowHamilton cycle, which is a contradiction.