If sound passes through material, vibration is produced. So are electromagnetic waves produced too? Sound means vibration of molecules and vibration produces electromagnetic waves. So, this means that sound produces electromagnetic waves directly. 
Is this possible?
 A: Physically vibrating atoms with a sound wave will technically accelerate the charges, but the radiation emitted will be very weak. However secondary effects from the sound wave are a different matter.
Indeed, sound waves in water can create light. This is called sonoluminescence. In the lab it's done with ultrasound passed through cavitation bubbles. It also happens naturally, with the pressure waves created by the rapid motion of the mantis shrimp's claw.
The exact mechanism is as yet unknown. One hypothesis is that the physical shock from the sound wave ionizes the particles of dissolved gases, which recombine and emit light.
A: A sound wave passing through a medium (e.g. air) indeed displaces molecules by a distance of a few nanometers. It seems reasonable that it should also displace the atoms, and thus electrons and protons in the process, which are charged particles and should radiate by Larmor's equation when undergoing acceleration.
Let us assume the sound frequency is on the order of kHz (which is in our audible range). Then molecules are accelerated by
$$a \approx \left(10^3\, \mathrm{Hz}\right)^2\times \left(10^{-9}\, \mathrm{m}\right) \approx 10^{-3}\, \mathrm{m/s}^2 $$
Then the predicted power of the radiation produced by Larmor's equation is ridiculously small
$$P = \frac{2}{3}\frac{q^2 a^3}{c^3} \sim 10^{-73}\, \mathrm{W} $$
Even if one multiplies this by the number of molecules of air in a $\mathrm{m}^3$, $N\approx 10^{25}$, this would never be detectable. Therefore, this effect may well exist, but it is absolutely negligible in all respects.
N.b. My answer focused on direct effects of the acceleration of air molecules due to a sound wave. As other answers mention correctly, there are interesting secondary effects of (especially large-amplitude) sound waves involving EM radiation. Among these are sonoluminescence and heating of the air by sound dissipation leading to increased thermal radiation.
A: Motions of neutral atoms do not radiate
@Cyclone's answer provides very useful insight, but is ultimately irrelevant. As long as the medium that is vibrating is neutral, no energy is radiated.  The EM waves emitted by the protons in the air interfere destructively with those emitted by the electrons and cancel out.
Put another way, the emitted radiation is proportional to the integral of the net current density. If positive and negative charges are moving the same, the net current density is 0.
In order to radiate, you would have to accelerate the atoms fast enough that they move faster than the electron cloud can catch up, causing a separation of the center of positive and negative charge (i.e. forming a dipole).
A: If the vibrating material is charged, then electromagnetic radiation will be produced.
There is a type of microphone called the capacitor or condenser microphone, in which the microphone is a capacitor whose capacitance varies with the acoustic vibrations.
https://en.wikipedia.org/wiki/Microphone#Condenser_microphone
The capacitance can be detected either by measuring the impedance to an RF signal, or by applying a high voltage direct current bias and measuring the current flow in and out of the capacitor as its capacitance changes. 
Normally the microphone is connected directly to the amplifier circuit, but it should be possible to couple it indirectly. In this case the electromagnetic radiation produced by acoustic vibration would be detectable.
If the vibrating material is magnetized, then electromagnetic radiation will be produced.
A typical electric guitar pickup consists of several magnets (usually one per string) surrounded by a coil of wire. The magnets induce magnetism in the string. The vibrations of the magnetized string are picked up by the coil of wire. Hence the electromagnic radiation produced by the string is detected.
However, in the general case where nothing is done to make the vibrating material an emitter of electromagnetic radiation, it will be difficult to detect (unless the vibration is so intense as to cause the vibrating material to heat up and emit radiation due to its increased temperature.)
https://en.wikipedia.org/wiki/Pickup_(music_technology)
A: I am deeply surprised by the neglect of the energy balance. In the previous answers. All sound is dispersed within a body. The energy difference between the incoming sound and the outgoing sound turns into heat.
The increase in temperature of the body is accompanied by an increase of electromagnetic radiation. So your guess is right, the body is out of its thermal equilibrium and responsible for this this the sound. 
A: If it passes through a piezoelectric material, it may generate a measurable voltage and current, some fraction of which will be radiated.  For most materials, however, there won't be a detectable effect.
A: The short answer is that some vibrations in materials can decay through photon emission, but that type is not excited by acoustic waves. Check out optical phonons.
