One problem is that you are trying to use a BJT (Q2) as a switch for signal path between the error amplifier and the output FET. I'm not really sure what you are trying to accomplish there, but I doubt that Q2 is ever going to do what you want.
When Q2 is off, there's no DC current path to set M1's gate potential, so the state of M1 is then undefined. With Q2 in any other state, M1's gate cannot fall below Q2's base (remember there's a PN junction between collector and base), and so M1's gate potential is severely constrained by whatever Q1's collector is doing.
The other mistake seems to be that you are modifying the output of the error amplifier, which will not change the overall output at all. All that does is alter how far the op-amp has to raise or lower its own output, to achieve the same 5V at M1's source.
In the circuit below I've added 2V to the output of the error amplifier, using source V1, but all that will happen is the error amplifier lowers its own output by 2V to compensate. The output is still 5V:
simulate this circuit – Schematic created using CircuitLab
That closed loop tries to maintain 5V at the output regardless of what stuff you place in the loop, or at the output. If you want the loop to output less, then ask it in a more appropriate way:
simulate this circuit
Here I am using Q1's collector to become grounded when \$V_{CUT}\$ goes high. When that happens, R3 and R5 form a potential divider, cutting the reference input potential by 80%, resulting in the output also being cut by 80%.
The point is, change what the closed loop produces at its output by changing its input, not by messing with the loop in the middle. That's not a hard-and-fast rule, of course you can change the feedback loop, and there might be a hundred reasons to do that, but I don't think this is one of them.
An analogy might be that you don't slow down a drill by grabbing hold of the shaft/chuck. By interfering with its closed loop like that, all that happens is motor torque rises in compensation, to maintain the same speed. You slow it by asking it in the usual way, which is by releasing the trigger a little.
In your case that would be to lower the voltage you apply at the non-inverting input of the error amplifier.
Since you insist on interrupting the loop to take control of M1's gate, you need to be very sure that U1's output can no longer have any influence:
simulate this circuit
When DISABLE goes high, M2 (which could also be a BJT) shorts M1's gate to ground, disabling output. Be aware though, that in this state U1 will do its best to correct the error (it is after all an error amplifier), and its output will go as high as it can, saturating against the positive supply. When the disable signal returns low, it will take a few microseconds for U1 to recover, during which M1's gate is way too high, and you'll probably see the output spike to well over 5V until U1 can settle back into a reasonable, stable state.
I repeat, I don't recommend this approach, where you throw a spanner in the loop. Control the desired output state (0V) by modifying the potential at the non-inverting input of U1. Set that (input) to 0V instead; so that U1 is always in control, never producing wild swings at its output.
