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I burned a circuit by applying reversed polarity voltage. I measured the LDO that is supplied with 5 V and should output 3.3 V, however it only outputs ~2 V.
Now what can I deduce from that? I would expect the semiconductor LDO to not output anything after being burned, however it does. Could this mean that some of the auxiliary components (capacitors) are blown?
LDO in question is a XC6201P332 (datasheet)
I measured the LDO that is supplied with 5V and should output 3.3V, however it only outputs ~2V.
I have seen voltage regulators which produced a different (out-of-specification) voltage, after a reverse voltage input. Therefore based on experience, your ~2 V output could mean that the regulator has been damaged.
I would expect the Semiconductor LDO to not output anything after being burned
Electronics does not always "completely fail"; there are many partial failures which can also occur. Therefore I would not expect the same as you.
Could this mean that some of the auxiliary components (capacitors) are blown?
Capacitor damage is also possible, especially if there were any polarised capacitors (e.g. tantalum, which may even catch fire under reverse voltage conditions) on the regulator's input.
You didn't supply a schematic (and ideally BOM or component ratings/types, if not on the schematic), so it is impossible to be more specific. Close-up photos of the relevant components may also be helpful, in case they show any signs of overheating.
In summary: The regulator could have been partially damaged, which would fit with your result. Any other polarised device (e.g. capacitor) which was exposed to the reversed voltage, might also have been damaged.
