Filling up of orbitals
You are drawing very nice $\mathrm{p}_x$, $\mathrm{p}_y$ and $\mathrm{p}_z$ orbitals onto that piece of paper. But if you take an isolated atom, which direction is $x$? Which is $y$? The coordinate system you’re seeing the atom in is just something we planted over its top because humans like things nice and organised. It is entirely arbitrary.

By the way, the direct calculations do not give $\mathrm{p}_x$ and $\mathrm{p}_y$ orbitals (strictly speaking not even $\mathrm{p}_z$. They give you $\mathrm{p}_0$, $\mathrm{p}_1$ and $\mathrm{p}_{-1}$ (the indices referring to the value of $m_l$, the ‘magnetical quantum number’). In spherical coordinates, which are used because it makes the calculations easier, only the $z$ axis is semi-defined by values $\theta = 0$. Because $\psi(\mathrm{p}_0) = f(r, \theta)$ (the wavefunction of one of the p-orbitals is independent of $\phi$), we say that this represents the $\mathrm{p}_z$ orbital. $\mathrm{p}_1$ and $\mathrm{p}_{-1}$ (the direct solutions of the Hamiltonian) cannot even be mapped to directions as they involve complex numbers; only via linear combinations can we transform them into two real orbitals pointing in an arbitrary $x$ and an arbitrary $z$ direction.

Things change if you introduce an external magnetic field. The atom’s $z$ axis will now align with the magnetic field in some way, giving your system one preferred direction. One, not three. $x$ and $y$ are still essentially undefined. However, in a magnetic field it would make a difference which orbitals the electrons populate.

You might also be interested in checking out the Jahn-Teller effect which occurs in transition metal chemistry to avoid filling some of all degenerate orbitals. In a nutshell it is energetically better for certain complexes to distort themself from a perfect octahedron to create distinct energy levels that can be fully populated, half populated or unpopulated.