Choosing an orientation of a cube is like picking two edges of the cube, one edge between the left and right vertices and one edge between the front and back vertices. But it also assigns an orientation to each of these edges, because putting red on the front and green on the back is not the same as putting red on the back and green on the front.
You need to assign an orientation to each of the four front-back edges, selecting one vertex to be the front vertex, and one to be the back vertex. Each of the four front vertices must be distinct, and each of the four back vertices must be distinct.
In your picture the edges are symmetric. You should put arrows on them. The arrows on the four front edges can point from the front vertex to the back vertex. Then the edges join up into paths.
Now suppose your choice of edges and arrows is actually a solution to the puzzle. Consider the four front-back edges. Start at one vertex, say Red. There must be precisely one front-back edge that comes out of this vertex, so you can follow it to its corresponding back vertex. Then (unless the new vertex is also Red) you can do it again. You can repeat this until you return to where you started. (This must happen; you never return to any other vertex first.) This defines a simple cycle in the graph. Now if you have not yet visited all four vertices you can pick another vertex to start at, and find another cycle. This will continue until you have a set of between one and four simple cycles that contain all four north-south edges and that together contain each vertex once. Let's call graph that is the union of these cycles “$FB$”.
Now we can do the same thing for the left-right edges, and generate another graph $LR$. Both $FB$ and $LR$ are subgraphs that have four edges each, one from each cube. They are disjoint. And each is a union of disjoint simple cycles.
Finding such $FB$ and $LR$ can often be done just by inspecting the complete graph of 12 edges. Once we have found such, we have a solution to the problem. (In general, for arbitrarily many cubes, it is a hard problem.)
Now let's look again at your graph:
Clearly we can take the red edges to be the $FB$ set and the blue edges to be the $LR$ set. But we also have to assign orientations to the edges, so that the red $FB$ subgraph is made of simple cycles and the same for the blue $LR$ subgraph. For the red subgraph we might assign the orientations so that the path goes from red to green to white to black to red, which corresponds to the partial solution:
But your proposed solution has:
This doesn't form a cycle, because the $G$ vertex has two edges coming out of it, and the $W$ vertex has two edges coming into it. So it is not in fact a solution to the puzzle.