It's momentum that causes the recoil, not kinetic energy directly.
Let's first calculate it for a bullet shot from a "slugthrower" gun. Bullets and their "muzzle velocities" vary widely, so I'll use some figures that I found for deer hunting. I'll depend on some hunting enthusiast to set me straight if need be.
Hunter Mass: 200 lbs = 90 kg
Gun Mass: 9 lbs = 4 kg
Bullet Mass: 150 grains = 10 grams
Velocity: 3000 fps = 900 m/s
Momentum: 9 kg*m/s
Gun recoil velocity: 2 m/s
Hunter recoil velocity: 0.1 m/s
I'm ignoring the combustion gases from the bullet cartridge's explosive powder, but that likely has a comparable effect.
In general,
(momentum) = (total massenergy) * (velocity)
where
(total massenergy) = (rest mass) + (kinetic energy)/c^{2},
Yes, Einstein's famous formula E = m^{2}
This rifle bullet's kinetic energy increases its total massenergy by about 5*10^{12}
Now let's calculate for a laser. I'll use the kinetic energy of that bullet as a reference; it is about 4000 joules. The momentum of that energy of light is very small, because light travels very fast: 1.4*10[sup]5/[sup] kg*m/s.
This is about 1.5*10^{6} times that bullet's momentum, so one will not feel any kick from a laser pulse with that energy.
The next question to ask is whether that will be enough to do any significant damage. A laser will have to vaporize some material in its target, and the departure of the evaporated material will kick back.
That energy is enough to vaporize about 1.8 grams of water, 0.37 grams of aluminum, or 0.64 grams of iron. The vaporized material would have a perpendicular velocity of about 500 m/s, giving 1, 0.2, and 0.3 kg*m/s momentum for each material  much less than the momentum of that bullet.
So one will likely need energies of about 200 kilojoules to produce a vaporization kick comparable to that bullet's momentum. But even that will produce less than about 10^{4} of the bullet's recoil momentum at the laser itself.
