I could see this being on/off control (a switcher) if there was a inductor in series with the FET. However since there isn't, this won't work as a switcher.
This topology looks like it is intended for linear operation. However, the capacitor slows down the system response so that the controller could easily oscillate under some conditions. This is not a good idea in this case. The way to fix it is to slow down the controller. This can be done by "compensating" the opamp if it has such control. If not, the right size cap between the opamp output and its - input will slow it down. For that to work, there will have to be a resistor between Vout and the - input. You can start with a 1 kΩ to 10 kΩ resistor and then use the capacitor value that is just a little above the minimum where the system doesn't oscillate at all operating points you care about.
There are more sophisticated ways to control power supplies, but what I described above is a quick fix that only adds two parts to the existing topology. The tradeoff is that is will slow the transient response. If you care about performance, then you need to start from the beginning and design a power supply the right way.
First observation: Decoupling capacitors ought to be connected at both the input and output legs of the regulator, as close as possible to the pins, connecting to the ground rail. Without those, most likely the regulator is going into oscillation at the output.
Notice that your observed output voltage is exactly half the expected output - an indication that the output is oscillating between 5 and 0 Volts. A DC voltmeter won't catch the oscillation, just the resultant average voltage. An oscilloscope trace will show it.
Second observation: Even though the regulator is rated for 1 Ampere, and the load is probably rated for much lower, this does not take into account the surge current the Raspberry Pi board needs at start-up. That surge is most likely forcing the regulator into over-current protection, and that'd be one of the probable causes for the oscillation.
The resistor on the other hand is a passive load, which will carry essentially a constant current both at start-up and later. Hence no surprise current surge to destabilize regulation.
Adding a suitably large decoupling capacitor at the output leg of the regulator will also help damp the surge load on the regulator, hence it is quite feasible that the regulator and your Pi will stabilize. If not, you may need to use a higher rated regulator to cope with the surge and the operating current needs of your arrangement.
Final observation: Some linear regulators need a basic minimum load current for stable regulation. This can be achieved through a load resistance next to (parallel to) the output decoupling capacitor - the resistance to use would be calculated to just barely draw the required minimum load current. Though this is probably not the problem in your circuit, it is useful to know, and address, based on information in the regulator's datasheet.
Best Answer
The power supply clearly has symmetrical positive and negative output rails, which only works if the two grounds are connected together. I suspect the grounds are actually connected together at the chassis, and the reason for keeping them separate elsewhere is to avoid a ground loop between the power amp and sensitive input circuitry.
Q214 drives the JFET which switches the input signal into the amp. The JFET turns off when negative voltage is applied its Gate, so the input is enabled when Q214 is off and its Collector voltage rises.
D201 rectifies AC from the transformer and produces a negative voltage on Q214's Base (more negative than the Emitter which is at B-), so the input is only enabled when mains power is present. This ensures that the amp will go silent as soon as it is switched off. R214 and C215 provide a delay between power on and input enabled, which may stop the amp from producing a 'thump' when it is turned on.