gate current direction during turn-ON of P-channel mosfet
No, also for P-channel MOSFETs it's charging them. The confusion probably stems from the voltage you see at the lower side, which goes to ground (or near it). But that voltage isn't important, a capacitor's charge is determined by the voltage across it:  

\$ Q = \Delta V \cdot C \$

So decreasing the gate voltage increases the gate-source voltage difference, which increases the charge of the capacitor.  

When you switch off T1, there's current flowing from +12 V through R2 to the gate to discharge it's capacitance.

edit re the update of your question dd. 2012-07-09(*)
Turning off means that you discharge the gate to +5 V, and this happens by current through R2 and D3. So you bypass R? but R2 is still the limiting factor. A solution would be to swap R2 and T1, so that there's more current/less resistance to discharge the gate than to charge it.  

\$ \$

(*) I'm using the ISO 8601 standard date format here. We have user from all over the world and for some 9/07 means 9 July, for others it's 7 September. ISO 8601 is unambiguous.