What would be the number of inequivalent $6$-colourings of the faces of a cube?

There are $6!=720$ different ways of colouring without taking rotations into consideration. There are $24$ different rotations, and for any colouring, no two rotations give identical result. Therefore these $720$ colourings may be divided into groups of $24$ essentially equal colourings. There must be $30$ such groups.

With your approach, you're choosing three pairs of oppositely placed colours, but you are overcounting because the same division into three pairs can be chosen in $6$ different orders, being counted erroneously as distinct by you while. At the same time, you consider one colouring and its mirror image non-distinct (swapping two opposite colours is erroneously seen as the same choice). This gives a total overcounting by a factor of $\frac62=3$.

In a circular permutation, arrangements which may be obtained from one another by rotating the cube are considered identical.

Place a colour on the bottom. It does not matter which. The top colour may be chosen in five ways. Place another colour facing you. Again, it does not matter which. The remaining three colours may be arranged relative to the colour facing you in $3!$ ways as we proceed clockwise around the cube. Hence, there are $5 \cdot 3! = 30$ admissible colourings.