With 1 meg from the gate of Q1 to GND, and SENS25% at 50k, Q1's Vgs will be 4.76 V, while with SENS25% at 600k, Vgs will be 3.125V.
That means that Q1 is almost certainly capable of being fully turned on - albeit somewhat slowly - by SENS25%, but R2 will prevent that from happening because the higher Vgs goes, the higher the current through R2 goes, which will cause the voltage dropped across it to increase, raising the voltage on the source, partially defeating the rise in Vgs.
A solution might be to wire your circuit like this:
Where, when Q1 is OFF, 5V across R1 R3 R4 will make Vout equal to a "1" into your Raspberry Pi, and when Q1 is ON, the drop across the LED and Q1's Rds(on) will make Vout equal to a "0".
If not, I think you'll need some extra circuitry or a different way to go about it.
This is not recommended. Sometimes MCU inputs specify that a certain amount of current can be injected through the protection diodes. You will have to check the MCU you eventually settle on. Select a pullup that will insure you stay under the maximum allowable current. But it would probably just be easier to add an additional NPN or N-channel (BSS138) to pull down the gate of the PMOS. Then there is no problem and you are free to choose any MCU, because it will never "see" 12 V.
Edit: I was only thinking about damage to the MCU. But @abdullah kahraman raises another equally important point. The diode current will also cause the PMOS to turn on. So this just plain won't work.