The answer is no.

Take the picture of a 5x11 rectangle and top it with a 6x11 made of 
2x2 and 3x3 tiles.  So the whole 11x11 can be split into 3 rectangles each 
tileable by two of the sizes. But if we cleave it we get an 11xA and an 11xB 
where A+B=11.

Using tiles of sizes axa and bxb we can tile rectangles of size 11xX with

X=6k if a=2 and b=3
X=10k if a=2 and b=5
x=15k if a=3 and b=5

This clearly does not enable us to find A and B with A+B=11.