Light scattering is a relatively complicated process.
In short, the beam will get wider and lose power, making it work less well.
Both of these reduce the intensity. If you like, think of intensity as pressure, as both intensity and pressure are inversely related to area. That is to say if two beams have the same power but one has a wider spread and thus larger area, that laser will do less 'damage'. In the same way, a bullet which hits with the same force but has a larger cross section will do less damage.
Now on to Scattering:
Unfortunately water is not really like crystals, which diffracts light into patterns. Water and most liquids are what we call amorphous. Here is a picture of Water (amorphous) on the left and Ice (crystalline) on the right:
Basically the atoms in an amorphous liquid cause the light to bounce every which way, very little lines up and it gets scattered.
The main process that occurs in light scattering is Mie Scattering, but a decent approximation is Rayleigh Scattering.
Ignore everything there except the second bracket.
- In Rayleigh Scattering, the attenuation (loss of power) is proportional to the inverse of the wavelength (lambda, which looks like an upside down Y) to the power of 4! That's right, have a short wavelength and the attentuation shoots up. If we double the wavelength we increase the attenuation by a factor of 16, if we triple it it goes up by a factor of 81!
That's all well and good, but what does it looks like? Well that depends on the size of the particles in the water. Unless the water is completely clear lots more scattering takes place.
Here is a comparison of scattering in clear water, and milky water.
It enters from the right, get's scattered in the murky liquid then passes straight through the clear water. The pattern on the far left has a much bigger area around it where the light has spread out.
Here is another one where we can see how the light is spread out in the 'turbid' fluid, and it doesn't go that far. This is the widening and the loss of power I talked about at the start.
To understand how the different wavelengths interact with water and other particles, here is some data:
You can see, major scattering occurs from the particles in the water but not the water itself (other factors do play into it, and the difference between air and water does matter over long distances in water).
I would liken this scattering effect to turning the laser rifle into a much weaker sawn off laser shotgun. Not as powerful but with more spread.
In short: yes it will work, just not as well.
[Edit: as requested from wikipedia:
Absorption plays a much greater role in clear water, and is dependant on what vibrational modes the EM wave excites in the molecule. See here for more of an explanation. TL;DR if it's anything but visible light it's going to struggle to pass more than a few cm through the water.