How to find if the unpaired electron is placed in hybrid orbital or pure orbital in odd electron species?
When discussing hybrid orbitals, they are justified by the experimentally observed or calculated/simulated molecular geometry. For instance, the $\ce{^.C(CH3)3}$ radical was found to be planar by ab initio calculations (http://www.tandfonline.com/doi/abs/10.1080/00268977600102711). Thus, the molecular shape is best approximated by $\ce{sp^2}$ hybridization and the unpaired electron is most likely to be found in a $\ce{p}$ orbital.

As a simple heuristic under the arguably outdated VSEPR theory, electronegative atoms want to be the furthest away from each other and other electrons as possible due to electric interactions. When considering the $\ce{^.CF3}$ radical, the fluorine atoms are highly electronegative and want to be the furthest they can be from the unpaired electron that holds negative charge. A planar $(\ce{sp^2})$ conformation would make the distance between the unpaired electron and each fluorine atom $90^\circ$. While a pyramidal $(\ce{sp^3})$ conformation would separate the two groups by $109^\circ$, which is much more favorable according to VSEPR theory. The later is what is observed under calculations (http://www.chem.mun.ca/courseinfo/c4420/handouts/handout-03a.pdf) with a $\ce{F-C-F}$ bond angle of $112^\circ$.

So, when considering hybridization problems, you must remember that hybridization fixes geometry. Thus if any plausible argument for a hybridization must be made, there should be experimental or computed evidence of the geometry. One cannot simply look at a molecule and know what the geometry could be with absolute certainty, but we can make a good guess based on simple heuristics, like those of VSEPR, molecular orbital diagrams, previous experience/similar molecules, etc.