Using discontinuous pressure spaces has the advantage that the solution is cellwise conservative (because you can test the equation $\nabla \cdot u=0$ with the characteristic function of each element). Thus, there is an advantage for using the $Q_k - P_{-(k-1)}$ element on quadrilateral/hexahedra. People do indeed use this element in practice.
On the other hand, this implies 3 pressure degrees per cell (in 2d) as opposed to roughly one per cell for the $Q_1$ element. So it's more expensive. But a practical observation is that it is not more accurate, but in fact often less accurate than just the regular Taylor-Hood element. The choice therefore comes to "better accuracy and cheaper" (Taylor-Hood) vs "locally conservative but more expensive" (discontinuous pressures).