Are the p orbitals of the biradical dioxo compound in the HOMO perpendicular to the plane?
I think you are asking what does the HOMO and LUMO of the 1,3-dioxabiradical look like.  

Let's start by using cyclopropane as a model compound.  In cyclopropane the $\ce{C-H}$ bonds are approximately  $\ce{sp^{2.46}}$ hybridized and the $\ce{C-C}$ bond is approximately $\ce{sp^{3.74}}$ hybridized (see here).  That's a lot of p-character in the $\ce{C-C}$ sigma bond. Dioxirane is likely similarly hybridized.  As soon as the 1,3-biradical forms (as soon as we break the $\ce{O-O}$ bond), the HOMO and LUMO will look like what is shown in the "initial" drawing below.  If the biradical lives long enough to undergo rotation about the $\ce{C-O}$ bond, then the HOMO and LUMO will look like what is shown in the "rotated" drawing.



These drawings just point out the conformational dependence of the HOMO and LUMO.  Note that the two lone pairs on each oxygen have not entered the discussion.  They are orthogonal to, and have no effect upon, the HOMO and LUMO orbitals.  

Imagine bringing two oxygen atoms closer to each other.  If we push them closer and the orbitals containing the odd electrons are pointing towards each other as in the "initial" geometry, we will form a sigma bond and we know what the HOMO and LUMO look like for a sigma bond (just like what is pictured in the "initial" geometry, only the orbitals are closer together).  If we push them closer and the odd electrons are pointing parallel to each other as in the "rotated" geometry, we will form a pi bond and we know what the HOMO and LUMO look like for a pi bond (just like what is pictured in the "rotated" geometry, only the orbitals are closer together). In the 1,3-biradical the carbon atom holds the 2 oxygen atoms together, they are closer than they would be at infinity, but further apart then they would be if there was a bond between them.  This tells us that the HOMO-LUMO energy separation is greater than it would be at infinity (0), but less than if they were closer together.

We could refine our model by noting that, according to Bent's rule, we would expect the orbitals containing the oxygen lone pairs to have a bit more s-character that the carbon $\ce{C-H}$ orbital in cyclopropane.  Therefore, in dioxirane and the corresponding 1,3-dioxabiradical, we would expect both the $\ce{O-O}$ bond and the orbital containing the single electron in the biradical to have even more p-character than $\ce{sp^{3.74}}$.

Again, however, to re-emphasize an earlier point, the lone pairs on oxygen have no bearing on our discussion.  That said, if we would like to speculate on the hybridization of the orbitals containing the lone pairs, we may do so.  Most texts tell us that the oxygen lone pairs in water, ethers and alcohols are in (roughly) $\ce{sp^3}$ orbitals.  However data from photoelectron spectroscopy experiments seems to suggest that one lone pair is in (roughly) a p-orbital while the other lone pair is in an $\ce{sp}$ orbital.