Divide the 8 vertices of the cube into two sets $A$ and $B$ such that each set contains 4 vertices, any two of which are diagonally adjacent across a face of the cube. We do casework based on the number of vertices of each color in set $A$. - Case 1: 4 black. Then all the vertices in $B$ must be black, for 1 possible coloring. - Case 2: 3 black, 1 white. Then there are 4 ways to assign the white vertex. The vertex in $B$ surrounded by the black vertices must also be black. Meanwhile, the three remaining vertices in $B$ may be any configuration except all black, for a total of $4\left(2^{3}-1\right)=28$ possible colorings. - Case 3: 2 black, 2 white. Then, there are 6 ways to assign the 2 white vertices. The 4 vertices of $B$ cannot all be the same color. Additionally, we cannot have 3 black vertices of $B$ surround a white vertex of $A$ with the other vertex of $B$ white, and vice-versa, so we have a total of $6\left(2^{4}-2-4\right)=60$ possible colorings. - Case 4: 1 black, 3 white. As in case 2, there are 28 possible colorings. - Case 5: 5 white. As in case 1, there is 1 possible coloring. So there is a total of $1+28+60+28+1=118$ possible colorings.