I think a counterexample can be obtained via [simplicial spheres](https://en.m.wikipedia.org/wiki/Simplicial_sphere). It is known that in higher dimensions, most simplicial spheres do not belong to a polytope. The dual of the face lattice of such a $d$-dimensional non-polytopal simplicial sphere is a $d$-regular "simple" poset in your sense, and cannot belong to a $d$-polytope (as otherwise, the original poset would too). However, it is not clear that the graph of such a poset does not belong to a $d$-polytope either.

Another approach that comes to my mind are [abstract polytopes](https://en.m.wikipedia.org/wiki/Abstract_polytope) which are defined purely in term of posets. Here, I have two explicit 3-dimensional examples from the class of [projective polyhedra](https://en.m.wikipedia.org/wiki/Projective_polyhedron):

 - the face structure of the [hemi cube](https://en.m.wikipedia.org/wiki/Hemicube_(geometry)) does not belong to a polyhedron, but its graph is $K_4$, which does belong to the tetrahedron.
 - the [hemi dodecahedron](https://en.m.wikipedia.org/wiki/Hemi-dodecahedron) cannot be realized as a polyhedron. Its graph is the [Petersen graph](https://en.m.wikipedia.org/wiki/Petersen_graph), which is not planar and hence cannot belong to a polyhedron either (in fact, it cannot belong to *any* convex polytope). Its poset can be construct quite easily from the face lattice of the dodecahedron by identifying antipodal faces.

Similar examples can be obtained from all centrally symmetric simple polytopes.