If there are no accepting paths or only one accepting path, it outputs zero. And If there are more than one accepting paths it outputs Yes. I am looking for a natural problem that requires this.
In a cubic graph, the number of Hamiltonian cycles containing a given edge is even. Therefore, if a cubic graph has any Hamiltonian cycles, it has at least two. (The same has been conjectured to be true for $k$-regular graphs with $k \ge 4$ as well; I don't know the status of that conjecture.) So deciding whether a cubic graph contains a Hamiltonian cycle (which is still NP-hard) seems to meet the criteria.
However, showing that this (or any other) problem is not in UP seems difficult, since you would have to show that the multiplicity of solutions can't be eliminated by some clever non-parsimonious reduction. It's not enough to show that the obvious NP machine that solves the problem has multiple accepting paths.
Consider the NP-complete problem, for $k>2$, of $k$-coloring a graph: Either the graph is not $k$ colorable, or it has at least $k!$ distinct $k$-colorings, simply due to permutation symmetry of the $k$ colors.
For $k=2$, the problem matches your specifications: 0 or 1 accepting paths (the latter of which will never occur) will not convince the verifier, while 2 solutions (or multiples thereof) will.