Do we consider the spatial orientation of a molecular orbital while assigning plane of symmetry?
The reason that orbitals are not considered in the symmetry of a molecule is that the orbitals cannot decrease the symmetry of the molecule. If we look at the canonical delocalized molecular orbitals of any molecule, everyone of them is symmetric or antisymmetric with respect to a symmetry operation of the molecule, with those symmetry operations determined only by the positions of the atoms, not the electron density.
Looking specifically at carbon dioxide, the simplistic picture of $\ce{CO2}$ as being like allene with lone pairs instead of the C-H bonds is not an accurate view of the electron density. If one looks at the delocalized pi-type orbitals, both span all three atoms in either the xz or yz planes (using the standard convention of the z axis being aligned with the molecular axis). Thus, both orbitals are symmetric with respect to the xy plane of symmetry of the molecule (which passes through the C).
Similarly, the lone pairs do not occupy separate sp2 orbitals that have specific alignment in a plane perpendicular to the p orbital involved in pi bonding. Instead, the create a combined electron density that can be deconvoluted into an sp-type orbital along the z-axis and partial contributions from the px and py orbitals, such that the total has no bias along the x or y axis.
If we were to look at the total electron density of $\ce{CO2}$, we would therefore see only a pill shape that does not vary as one rotates around the z-axis.
To visualize this, I recommend looking at all of the calculated delocalized molecular orbitals of $\ce{CO2}$. You can do this on the molcalc.org website if you do not have other software for that purpose.