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On all the IO pins is a diode going from the pin to the Vdd rail. It starts to forward-bias when Vio>(Vdd+0.3v). I don't know if it's actually a schottky or std silicon diode. The spec sheet says 0.3v which implies schottky, but that may be to a very low criteria for current. Anyhow, that means that putting 5v on the IO pin results in 4.7v on Vdd rail and the Vdd power pin with no power hooked up.
The chip may start running. However, that diode is not a power diode. It was not meant to bear current. If the chip has no significant output loads (like LEDs), then the diode's rating may be ok. If you have power loads- or, get this- say the Vdd trace was broken before it reaches the regulator on the PCB, but that Vdd trace reaches other loads on the board. That Vdd trace and the other loads may be energized THROUGH the IO pin.
The same situation can result if the ground pin is disconnected.
If the current's high enough, then the diode can fry. How much current that diode can take is not really clear, I've asked that of chip makers before and they said "uhh... we don't specify, but 'not much' ". The 12F-18F series are tough though. I doubt you could blow them with any likely setup. There are newer 18x parts designed with a different silicon process solely for low voltage and I can't attest to those one way or the other.
Note this may ruin your reset strategy. Say you've got a large capacitor or battery or whatever on an IO, some independent voltage source. If you remove power, the device will remain powered through that pin. Restore power and it won't undergo a reset. It's potentially possible that the voltage will drop into the "brownout" range where it scrambles the RAM states, but restoring power is not enough of a voltage gain to guarantee a reset on MCLR, resulting in a powered but nonfunctioning state.