You obviously have to consider the current through the resistors (and make sure you choose inputs with protection diodes!), but it's a perfectly valid method, and is approved by MicroChip (to the extent it features in many of their applications notes).
Surely, an LED to indicate power on will be sufficient to absorb the 10mA that will be supplied via the input diodes. If you don't have an LED then the max voltage of 7.5V on Vdd (=8.2 on each pin) would raise the current consumption to at least 10mA. That is assuming no output pin has any load on it.
The problem comes into play if you try to make an efficient circuit with no power LED and especially if you try to use SLEEP mode which will lower the PIC current to near 0.