Below you see an example of a bicubic graph consisting of faces with degree $4$ and $6$, which makes up the set of graphs of my interest and is a subset of the so called Barnette graphs.
Since the graph is planar it has a 4-coloring of the faces and therefore a 3-coloring of the edges. If you color the edges along the Hamilton Cylce (HC) given by the increasing labels with the colors 1 and 2 (color 0 is left for the non HC edges).
Using this 3-coloring you can assign an orientation to every vertex. Just look at the graph and check how the colors 0,1 and 2 result in an orientation. For the example above you get: $$ LRLL|RLRR|LRLL|RLRR|LLRR $$
E.g. at vertex 2, the edge (1,2) has color 1, edge (2,3) has color 2 and edge (2,9) has color 0. So the vertex has a right orientation, i.e. $R$.
The first observation from the examples we have so far (the cube and another 12 vertex graph) is that the orientations occur balanced, i.e. you get the same number of left and right oriented vertices. (Proven below...)
But if you now go along the HC, i.e. you start from 1 and go to 2 you have to turn left, which is against the orientation of the vertex, which is right. I write $0$ for wrong and $1$ for correct orientation. What you now get is, at least stunning to me: $$ 01010101010101010101 $$ How to prove that?
I also checked an alternative HC for the graph given above and some simpler graphs with the same result.