Yes. Because Eulerian circuits are a subset of Eulerian trails, all Eulerian circuits must be traversable since, by definition, a Eulerian trail is traversable.
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"The rule to find whether a network is traversable or not is by looking at points called nodes. Nodes are places where two or more lines meet. On these networks, the nodes are clearly shown by the black points in the diagrams. Now you are probably wondering what this has to do with the network being traversable or not. The node either would have an odd or even number of lines connected to it. Do not count the nodes with an even number of lines connected to it. Count the number of nodes with an odd number of lines connected to it. If there are no odd nodes or if there are two odd nodes, that means that the network it traversable. Networks with only two odd nodes are in a traversable path and networks with no odd nodes are in a traversable circuit."
It is traversable if there is an even number of edges at each vertex, or at every vertex except two. In the latter case the traverse must start at one of the "odd vertices" and finish at the other.
The definition of an Eulerian path is a path in a graph which visits each edge exactly once. Intuitively, think of tracing the path with a pencil without lifting the pencil's edge from the page. One definition of an Eulerian graph is that every vertex has an even degree. You can check this by counting the degrees. Please see the related link for details.
Each one of a cube's vertices has a valency of 3. The graph of its edges is therefore non-Eulerian and so it is not possible to have a cube route.
Traverse means "to cross". It's origin is the French word "traverser" which means "to cross".