Edging out a coloring

Let the colours be $1,2,3,4.$ Notice that there are 6 ways to cyclically order the numbers $1,2,3,4,$ so each ordering must occur on one of the six faces of the cube. Consider the ordering $1,2,3,4.$ There are 6 choices for which face this occurs on and 4 choices for which edge on that face has the colour 1. We label the other edges as in the figure of the cube below:

The edge $a$ can be either 2 or 3 so as not to violate condition $A,$ similarly the edge $b$ can be either 3 or 4. If neither edge is 3, then we can not complete the top face without violating one of the two conditions. We also cannot have both being 3, since this violates $A$ . If $a = 3$ , then $b = 4$ , $c = 1$ , $d = 2$ and the other edges are uniquely determined by rule $A$ and all faces will satisfy rule $B.$ Similarly, if $b = 3,$ then $c = 4$ , $d = 1$ , $a = 2$ and the rest of the edges are uniquely determined. This gives two colourings of the cube for each of the choices for our first face, for a total of $2 \times 24 = 48$ colourings.