The others are pretty much correct… the problem is not that you’ll burn up the 12V power supply; it’s that the voltage regulator inside the phone will run hotter.
To illustrate, here’s an example.
Let’s say there’s a 6V linear regulator inside the phone. Let’s also say the maximum current the phone requires is 500 mA (i.e. the load requires 6V at 500 mA).
If the input voltage is 9 VDC, the regulator will dissipate a maximum power of 1.5 watts. Most TO 220-packaged devices (which includes most through-hole regulators) have a junction-to-ambient thermal resistance of around 40 °C/W without a heat sink. Assuming the ambient temperature is 25 °C, the regulator’s junction will be 85 °C. This is below the 125 °C “safe” limit for such devices.
But let’s say you boost the input voltage to 12 VDC. Because the load still requires 6V at 500 mA, the regulator will now dissipate a maximum power of 3 watts, and the regulator’s junction will be 145 °C. This is above the 125 °C “safe” limit for such devices. (It could even worse than that. Though the 40 °C/W rating assumes no heat sink, it also assumes the package is standing “upright,” thereby exposing all surfaces to the air. If the back surface of the package is screwed to the PCB as is commonly done, the effective junction-to-ambient thermal resistance goes up.)
The saving grace is that most regulators have thermal overload protection, which means you probably won’t permanently damage the regulator, at least in the short run. But in order to shut down, the regulator must exceed 125 °C for a certain amount of time. Do this enough times, and you will decrease the life of the regulator.
Solution: Get a 9V power supply.
Mouser Electronics sells them very cheap. Plus they do not have a minimum $ amount for an order: