To get rid of duplicate counts, put the cube on a table, with colour $1$ down. (i) If colour $2$ is not on the "up" side, rotate the cube until colour $2$ is facing you. (ii) If colour $2$ is on the up side, then rotate the cube (keeping colour $1$ down) so that colour $3$ is facing you.
Counting the possibilities in (i) should now be straightforward, as should counting the possibilities in (ii).
Alternately, there are $5$ possibilities for the "up" side. For each possibility, rotate the cube so that the smallest numbered remaining colour is facing you. Now the rest of the counting is not difficult.
Added: There are more general group-theoretic approaches (Polya counting), but it does not seem reasonable to bring them out for this fairly "small" problem.
For situations in which colours may be duplicated, a similar approach will work. However, because the possible symmetries are more complicated, Polya counting becomes more reasonable. But one can also break things down into cases, by examining separately $1$ colour, $2$, $3$, $4$, $5$, and $6$.