Ok, I'ma call it.
There's no way to get more than 4 of every colour. Also, there is no simple way to prove this.
This is the most efficient way to do it:
There's room for all kinds of shenanigans: You can add
* a white queen to e1 or a2
* a red queen to g2 or h7, and
* a black queen to e7, or
* any queen to g7, if you move a1 to a2 first
Annoyingly, adding any two colours always excludes every option of adding the third colour, no matter how much you shuffle the pieces around.
This way you can get
any two colours to 5, but not all. Also, you could get white to 5 (e1) and red to 6 (g2, g7), but black still stays at 4, so you get a 4-5-6 solution.
In addition to all that:
There's so much wiggle room in the above diagram, and you can get so very very close to a 5-5-5, that any simple impossibility proof (like "there aren't enough diagonals on the chess board") is not going to work.
This all is a result of feeding this problem into a highly sophisticated self learning neural network *, making it start from random (and later self-selected) positions, where every improvement path always led to this position, or one of its descendants, showing that this position is at least a local optimum.
Now all that is needed is
a) a brute force solution proving that this is indeed the optimum, or
b) a 5-5-5 solution, or a simple proof of its impossibility.
If anyone can provide case b, I'll happily buy that person a beer, after a solid stint of banging my head against a wall.