I can only provide an answer from an experimental quantum optics angle (i.e. low energy limit of QED):
Once you have defined a particular mode of the electromagnetic field (for example in a cavity, or any kind of spatio-temporal mode) and applied the usual "canonical quantisation" procedure, you are looking at a more or less large Hilbert space (formally equivalent to a more or less large collection of harmonic oscillators). The "vacuum state" is then just the state that has no excitation, i.e. in the number basis you would just write $|0\rangle$, in case of a single mode.
Experimentally, once you are in a dark room, the vacuum state is a very good approximation to the "true quantum state" of almost any modes that can be defined in the optical regime, as the vast majority of modes contain no photons (as you can even guess from your visual impression). This is even true in a bright room :D, there is just such a ginormous number of ways the electromagnetic field can oscillate.
If you go to lower energies like microwave regime (or hotter environments), the approximation is not so good anymore. In thermal equilibrium the field would be best described by a thermal state, where you as the experimenter have minimal information about the quantum state, you just know one parameter of the radiation field, and that is its temperature (determining the mean number of quanta in a particular mode).
So if you wanted to experimentally "prepare" the radiation field (or a single mode of it) in the vacuum state $|0\rangle$, you don't have to take crazy measures at optical frequencies, you get the vacuum state for free (which is a pure quantum state! This is indeed very beneficial for all of quantum optics). But at next order your main enemy is basically black body radiation of the environment. So an option to improve things would be to literally cool the cavity walls with a fridge.
So it has almost nothing to do with somebody evacuating a chamber of air.