Yes! The Georges graph has this property. This graph is the smallest known counterexample to the Tutte conjecture that any 3-connected bipartite 3-regular graph is Hamiltonian. It has 50 vertices and 75 edges.
I verified that every vertex is the initial vertex of a Hamiltonian path with some Sage code:
def homtrac_orbits(g):
orbits = g.automorphism_group(return_group=False, orbits=True)
return all(g.hamiltonian_path(o[0]) is not None for o in orbits)
def report(G):
print G.name() + ':', homtrac_orbits(G), G.is_hamiltonian(), G.is_bipartite()
report(Graph(">>graph6<<qO`a`_WO?O?_?_?OO???K?@O?B_?@??__?_I?A?_?_????@?A?@???\
O????g???`????C????????A?????O????C??????????K?????K?????S?????K?????\
B????OO?????__??????C??????C??????A?????C????????D_??????I???????[\
??????AG??????GA_", name="Georges"))
You can see it on SageCell but it takes some time to run. It prints "Georges: True False True" when complete.
The Ellingham-Horton graphs on 54 and 78 vertices are also homogeneously traceable but non-Hamiltonian. The 54-graph can be checked with
report(graphs.EllinghamHorton54Graph())
(included in the above SageCell). I also checked the 78-graph but this is too slow to include in the SageCell.
It seems reasonable to guess that the other known counterexamples to Tutte's conjecture—namely the Owens graph, the Horton 92-graph, and the Horton graph—are also homogeneously traceable, but checking in Sage is too slow for those to verify that way.
Recall that an Eulerian path exists iff there are exactly zero or two odd vertices. Since $v_0$, $v_2$, $v_4$, and $v_5$ have odd degree, there is no Eulerian path in the first graph.
It is clear from inspection that the first graph admits a Hamiltonian path but no Hamiltonian cycle (since $\operatorname{deg}v_0=1$).
The other two graphs posted each have exactly two odd vertices, and so admit an Eulerian path but not an Eulerian circuit.
Best Answer
I'm not sure if all graph theory books treat degenerate cases the same way, but Diestel's Graph Theory, at least, allows a path to have length $0$, i.e., to consist of a single vertex with no edges. If a graph consists of a single vertex $v$, then the path consisting of $v$ is vacuously Eulerian. It is also a Hamiltonian path, since it contains all of the vertices of the graph.