The sound intensity formula (I=ξ²ω²cρ) tells us that the volume of your alarm would drop in linear proportion to the decrease in atmospheric density. How this will sound will depend a lot on how you are depressuring the airlock.
The pitch comes from your medium so your alarm's pitch should remain unaffected because the gas mixture would not change.
Lastly, undistorted sound cannot be carried by more pressure than is available. At 1 atmosphere, we can perceive clear sound at volumes up to about 190bd. As the atmosphere dissipates, the alarm will begin to distort because sound is only clear when it's peak compression does not exceed twice the density of the medium. This distortion will start to take place sooner, and be more pronounced, the louder your alarm is.
These effects will be most noticeable if you are wearing a helmet or inside of a cockpit where the actual air pressure you are in is not changing. If your ears are exposed, the sound will much more quickly begin to sound distant and distorted as the air pressure in your ear drums exceeds the atmosphere around you. In a slow depressurization, your eardrums will 'pop' equalizing pressure several times as you loose pressure. In a rapid decompression, your eardrums could potentially rupture which would be both painful and deafening.
Some sound will also reach you traveling through physical contact with the hanger as TheLuckless pointed out, but there are so many unknown variables there I would not begin to be able to hypothesize what that would sound like exactly other than still being able to still hear a faint and badly distorted alarm as you approach vacuum that would disappear when you leave physical contact with the hanger if you have not already done so.