To expand on Xcheckr's answer:
The full equation for a single-frequency traveling wave is
$$f(x,t) = A \sin(2\pi ft - \frac{2\pi}{\lambda}x).$$
where $f$ is the frequency, $t$ is time, $\lambda$ is the wavelength, $A$ is the amplitude, and $x$ is position. This is often written as
$$f(x,t) = A \sin(\omega t - kx)$$
with $\omega = 2\pi f$ and $k = \frac{2\pi}{\lambda}$. If you look at a single point in space (hold $x$ constant), you see that the signal oscillates up and down in time. If you freeze time, (hold $t$ constant), you see the signal oscillates up and down as you move along it in space. If you pick a point on the wave and follow it as time goes forward (hold $f$ constant and let $t$ increase), you have to move in the positive $x$ direction to keep up with the point on the wave.
This only describes a wave of a single frequency. In general, anything of the form
$$f(x,t) = w(\omega t - kx),$$
where $w$ is any function, describes a traveling wave.
Sinusoids turn up very often because the vibrating sources of the disturbances that give rise to sound waves are often well-described by
$$\frac{\partial^2 s}{\partial t^2} = -a^2 s.$$
In this case, $s$ is the distance from some equilibrium position and $a$ is some constant. This describes the motion of a mass on a spring, which is a good model for guitar strings, speaker cones, drum membranes, saxophone reeds, vocal cords, and on and on. The general solution to that equation is
$$s(t) = A\cos(a t) + B\sin(a t).$$
In this equation, one can see that $a$ is the frequency $\omega$ in the traveling wave equations by setting $x$ to a constant value (since the source isn't moving (unless you want to consider Doppler effects)).
For objects more complicated than a mass on a spring, there are multiple $a$ values, so that object can vibrate at multiple frequencies at the same time (think harmonics on a guitar). Figuring out the contributions of each of these frequencies is the purpose of a Fourier transform.
Background
The way to reduce the sound is to reduce the pressure change across the shock, which can be done using an oblique shock.
In this case, the angle between the shock surface and the incident flow speed, $\beta$, enters into the pressure ratio for a hydrodynamic shock as:
$$
\frac{P_{2}}{P_{1}} = 1 + \frac{ 2 \ \gamma }{ \gamma + 1 } \left( M_{1}^{2} \ \sin^{2} \beta - 1 \right) \tag{1}
$$
where the subscripts $1$ and $2$ correspond to the upstream and downstream regions, respectively, $\gamma$ is the ratio of specific heats, $M_{j}$ is the Mach number in the $j$ region, and $P_{j}$ is the average pressure in the $j$ region.
As you can see, in the limit as $\beta \rightarrow 90^{\circ}$, we approach the standard expected limit for a hydrodynamic shock. This is equivalent to looking at a shock produced by a piston/driver with a planar surface orthogonal to the incident flow.
The magnitude of the pressure ratio, or the overpressure, defines the strength of the sound wave of the sonic boom at the source. If we measure this in decibels, then we can see from the description at https://physics.stackexchange.com/a/266046/59023 that the intensity observed goes as:
$$
L_{r}\left( r \right) = L_{i,src} + 20 \ \log_{10} \left( \frac{ 1 }{ r } \right) \tag{2}
$$
where $L_{i,src}$ is the intensity level at the source and $r$ is the distance from the source. This is one of the reasons why one solution proposed is to fly a supersonic jet at higher altitudes (other reasons are outlined in the above linked answer relating to sound attenuation at altitude).
Answers
Is it still a wave?
Well, that's a little trick since a sonic boom is what your ear registers when a supersonic object passes by. It is the result of a shock wave, which is inappropriately named as I discussed at https://physics.stackexchange.com/a/136596/59023.
Furthermore, if it is a wave, why does it make a boom sound and does it have a decibel number?
The boom is a discontinuous pressure pulse (well, technically it's two, an over- and underpressure pulse). Yes, one can estimate the decibel level as I described above in Equation 2.
Can we measure the decibels?
Yes.
All of this leads me to the question, is it possible to counteract, or cancel, the noise of a sonic boom?
Cancel, probably not. Reduce greatly, probably yes. One way is to change the shape of the nose cone of the jet, which alters the shock geometry (as I discussed above). For certain geometries, one can get a reduced pressure ratio (e.g., Equation 1) and thus, a smaller overpressure sound wave resulting in a "quieter" sonic boom.
Reference
The phase diagram image was taken from Wikipedia, courtesy of EMBaero - Own work, CC BY-SA 3.0, https://en.wikipedia.org/wiki/File:Obliqueshock.PNG
Best Answer
The speed of sound increases with increasing pressure. Assuming ideal behaviour the relationship is:
$$ v = \sqrt{\gamma\frac{P}{\rho}} $$
or equivalently:
$$ v = \sqrt{\frac{\gamma RT}{M}} $$
where $M$ is the molar mass.
In a gun barrel just after the charge has gone off the gas is under very high pressure and very hot, so the speed of sound is much higher than under ambient conditions.