Is it true to say that with 10v at the cathode, the tube is biased incorrectly and is therefore likely to be in full cut-off?
No. The grid 'may' be +20V and the tube would be hard on. IOW, you must also know the grid voltage to determine how the tube is biased. Bias voltage is really the difference of voltage between the grid and cathode. You must know both. That's why it's important to post grid voltages too, even if they are zero.
Or in other words, the grid is -10v in relation to the cathode so no current can flow to the plate? (and the guitar's tiny voltage isn't enough to get anything to move).
That's a better statement since you're now talking about the voltage between the grid and cathode.
Specifically, for your situation, 10v on the cathode indicates that there is current flowing thru the tube. For your circuit, there is no other way for 10v to be on the cathode. (Unless something is wrong!) But the high voltage on the plate indicates the tube is not conducting. So, which to believe? It really can't be both ways, can it? Well, it can if something is wrong. You gotta look for something that can cause both symptoms.
Think about this... 10v across that 600Ω cathode resistor means 16ma is flowing. That's way too much for that 12AX7, but don't think about that right now. That 16ma is also flowing thru that 100KΩ plate resistor and that means there should be 1600v dropped across the plate resistor! Can't be. But what if the cathode resistor had increased to 100K. Then, 10 volts would mean that only 0.1ma was flowing. There would only be 10 volts dropped across the 100K plate resistor too, so the plate voltage would be very high.
So, an increase in cathode resistance could cause both the cathode voltage and plate voltage to be high. That's why I wanted you to check that cathode resistance.