Geometries of methyl and silyl radicals
As is currently being discussed in this question, I'm not convinced that the methyl radical is planar, but even if not planar, it is certainly very close to planarity.  It is also certainly less pyramidal than the $\ce{.SiH3}$ radical as you state.

The $\ce{.SiH3}$ radical is more pyramidal than the $\ce{.CH3}$ radical due to differences in electronegativity between $\ce{Si}$ (1.9) and $\ce{C}$ (2.5) and hydrogen (2.1) and how these differences play into Bent's rule.

In a nutshell, Bent's rule states that, "atomic s character concentrates in orbitals directed toward electropositive substituents" (ref).

Since hydrogen is more electronegative than silicon it will draw more of the electron denisty in a $\ce{Si-H}$ bond towards its end of the bond (the reverse of what happens in the case of a $\ce{C-H}$ bond where the carbon is more electronegative than hydrogen) decreasing the electron density in the silicon atomic orbitals involved in the bond.

If there is less electron density in the silicon atomic orbital used in the $\ce{Si-H}$ bond, then according to Bent's rule, the silicon atom will hybridize so that less s-character is used in these atomic orbitals that now have a lower electron density, and will instead use that s-character in the orbital containing the single electron and stabilize it.  Increasing the p-character in the $\ce{Si-H}$ bonds and increasing the s-character in the orbital holding the single electron will tend to push the molecule towards an $\ce{sp^3}$-like, pyramidal geometry.

The reverse is true in the case of carbon.  Due to electronegativities, there will be more electron density towards the carbon side of the $\ce{C-H}$ bond (compared to the $\ce{Si-H}$ bond).  Due to this increased electron density around carbon, it is energetically less favorable to remove s-character from the carbon atomic orbitals involved in the $\ce{C-H}$ bonds (and destabilize this increased electron density) and transfer it to the orbital containing the single electron.  As a result the carbon atomic orbitals involved in the $\ce{C-H}$ bonds will have higher s-character compared to the analogous silicon case and $\ce{.CH3}$ will be less pyramidal.