It's not a disturbance, the liquid isn't supercooled in this case. It's right about at 0 degrees, though.
It isn't the pressure drop directly, because you can give an upper-bound estimate to how much cooling the pressure drop directly does based on the observation that the pressure is not more than a few atmospheres. That means that space the gas in the bottle gets multiplied by a factor of, say 10 (that's too big), and (kT times) the entropy gain per atom is
$kT\Delta S = kT log(10)$
and kT is 1/30 eV, so you get at best 1/10 eV per gas atom. The density of gas on top is 1/300 the density of water, and the maximum total cooling from expanding that gas is negligible per atom of liquid. There is no significant cooling and heating in compressing a liquid, since essentially no work is done in the process, so the pressure drop does nothing to the liquid.
But the pressure drop makes the soda supersaturated with respect to its dissolved CO2, and this CO2 is outgassing from a liquid state to a gas state, and this is a huge gain in entropy. By outgassing the dissolved CO2 into bubbles, each CO2 atom gets a 300-fold increase of available volume to roam, and this can cool the liquid by
$kT log(300) \approx {1\over 4} eV $
So you get $.25 eV$ per released CO2 molecule, which is comperable to the binding energy of a water molecule. So each outgassed CO2 freezes order 1 H2O, and this freezes a network of filaments of water around the outgassed bubbles, making a slush.