Short Answer: Using the mean of the square (or the square of the RMS) represents the energy in the acoustical signal.
Longer Answer:
Sound waves that you hear are made of vibrations in the air. There are many effects that coincide with this vibration, but lets focus on two: the air moves and has a velocity, and the air pressure changes.
Air that moves has kinetic energy, or energy stored up in motion. The velocity of the air is actually very complicated due to the fact that it is made up of many molecules bouncing around, but when analyzing sound waves it is usually fine to "smooth out" the air by treating it like a continuous fluid. Then, one can think of looking at just a little patch of the air and measuring how fast it moves. This is called the particle velocity (technically in a Lagrangian sense, but don't worry about that). The kinetic energy, as you may know, is related to the mass of an item times its velocity squared. The mass of this little packet of air is just the density of the air times the volume of the packet. We are imagining a very small chunk of air that doesn't have a specific volume, so we will just divide the kinetic energy by the volume to get rid of it and say we are looking at the kinetic energy density instead. Thus, we may write the kinetic energy density as
$$ E_k = \frac{1}{2}\rho v^2, $$
where $\rho$ is the mass density and $v$ is the particle velocity.
There is another form of energy that is important for acoustics, which is potential energy. Think of a spring: if you pull it, it stores energy that can be returned once the spring is released. Air acts like a spring as well. If you pull it apart the pressure of the air around it pushes it back to its original state. Potential energy for a spring takes the standard form
$$ E_s = \frac{1}{2}k x^2, $$
where $k$ is a spring constant and $x$ is the distance that you move the spring. However, since the force of a spring $F$ is $F=-kx$ and can be rearranged to give $x=-F/k$, the potential energy could just as easily be written as
$$ E_s = \frac{1}{2k} F^2. $$
The force on air is the pressure (times the area), and the spring constant is related to the stiffness of the air, or the bulk modulus $\rho c^2$, where $c$ is the speed of sound. Thus, without actually proving it, we may write the potential energy density for sound in air as
$$ E_p = \frac{1}{2}\frac{p^2}{\rho c^2}, $$
where $p$ is the acoustic pressure. The total energy density may then be written as
$$ E_T = \frac{1}{2}\rho v^2 + \frac{1}{2}\frac{p^2}{\rho c^2}. $$
From these equations you can see that the square of the pressure and particle velocity are associated with the energy of the wave. But, we can explain further. Without proving it, I know that for a sound wave that is just propagating in one direction (no echoes or spreading) we may write
$$ v = \frac{p}{\rho c}. $$
(This relation is called the impedance relationship.) For this rather common case we may then eliminate velocity from the total energy density and obtain
$$ E_T = \frac{p^2}{\rho c^2}. $$
Now, the acoustic pressure oscillates, as you said. That means that the total energy density of the air at a point also oscillates, bouncing from zero to values proportional to the sound amplitude over time. Perhaps it would be most helpful if we were to average out the energy. Since the oscillations happen so fast anyway, this is kind of like smoothing out the sound, just like we smoothed out the air molecules into a fluid. We may then write the time averaged total energy density as
$$ \langle E_T\rangle = \lim_{T\rightarrow\infty}\frac{1}{T}\int_0^T \frac{p^2}{\rho c^2}. $$
Since $\rho$ and $c$ are constants (to the level of approximation for sound waves, anyway), this just leads to
$$ \langle E_T\rangle = \frac{1}{\rho c^2}\langle p^2\rangle = \frac{p_{\text{rms}}^2}{\rho c^2}. $$
Thus, we may see that the time averaged total energy density of the sound wave is proportional to the mean squared pressure. We could also have written it as being proportional to the mean squared particle velocity, acoustic density, or acoustic temperature. In the end, for sound (and many other physical systems) the energy is related to the square of some oscillating quantity, and so we average over that and wind up with RMS quantities.
To just round out the discussion, I will comment on the use of decibels. People noticed that humans respond to sound roughly logarithmically compared to the amplitude, and so they started to take the logarithm of the energy. However, decibels are a relative measure, and so they had to pick some reference value. For air, they chose an amplitude (or RMS value) of 20 $\mu$Pa. Thus, the sound pressure level (SPL; not the loudness) is defined as
$$ \text{SPL} = 10\log_{10}\left(\frac{p_\text{rms}^2}{(20~\mu\text{Pa})^2}\right). $$
From the properties of logarithms, we can bring the square down and we obtain an expression familiar to just about every acoustician:
$$ \text{SPL} = 20\log_{10}\left(\frac{p_\text{rms}}{20~\mu\text{Pa}}\right). $$
Loudness depends on the specifics of how humans hear different frequencies, and so you get more complicated expressions for loudness, but they are based on the SPL formulation as well.
