Let $T_0, \dots, T_n$ be a collection of spherical triangles that form a triangulation of the $2$-sphere.
Can we enumerate the triangles in such a manner that $T_{i-1}$ and $T_{i}$ have exactly one common edge for all $1 \leq i \leq n$? Can we enumerate them such that $T_n$ and $T_0$ have one common edge?
This question can be seen through the lense of graph theory: let $V$ be a graph whose vertices are the triangles, and let $E$ be the set of edges (quite literally). The question is whether the graph G=(V,E) has a Hamiltonian path or a Hamiltonian tour.
Nope!
Finding a Hamiltonian cycle is almost equivalent to Tait's Hamiltonian Graph Conjecture that every cubic polyhedral (i.e. 3-connected planar) graph is Hamiltonian. That is because every cubic polyhedral graph gives you a convex polyhedron with all 3-valent vertices, whose dual is a polyhedron with all triangular faces, which induces a triangulation of the 2-sphere.
The counterexamples to Tait's conjecture thus give you triangulations of the 2-sphere with no Hamiltonian cycle among the triangles.
For Hamiltonian paths, the question is whether every 3-connected planar cubic graph is traceable. I didn't find anyone discussing merely untraceable graphs, but there are several papers looking more specifically for hypotraceable graphs (which have no Hamiltonian path, but removing any vertex leaves a graph with a Hamiltonian path).
Carsten Thomassen's 1981 paper "Planar Cubic Hypohamiltonian and Hypotraceable Graphs" (DOI 10.1016/0095-8956(81)90089-7) has
Investigating the relevant construction from his 1976 paper "Planar and infinite hypohamiltonian and hypotraceable graphs" (DOI 10.1016/0012-365X(76)90071-6), we find that it yields 3-connected graphs, so these are polyhedral and give rise to untraceable 2-sphere triangulations.
The smallest of these hypotraceable graphs has 460 vertices. "On cubic planar hypohamiltonian and hypotraceable graphs" (M. Araya, G. Wiener, 2011) finds smaller examples, in