Is momentum perfectly conserved at the particle level given the Heisenberg uncertainty principle? Discussions of conservation of momentum frequently use the metaphor of two billiard balls colliding. My impression is that this is not valid at the quantum scale - an illustration of the particles' trajectories should show the outgoing vectors with some uncertainty. Perhaps the total energy could still be conserved if the two particles were entangled in such a way that the imprecision of one particle's trajectory was balanced by the second particle's trajectory. Even if that was the case, I am not clear that the net vector would be as expected, which would therefore mean the momentum was not conserved.
An alternate way of viewing the problem: at the moment when two particles collide, the position is known very precisely (since they had to hit each other at the same place and time). Since momentum is complementary to position, this means the momentum has maximum uncertainty at that instant. While the momentum in a single collision may be perfectly conserved, perhaps the momentum being conserved is somewhat probabilistic such that over billions of interactions of billions of molecules (as the original force propagates) the original net momentum is not conserved.
I tried to look for answers to this question and here are some relevant ones. They seem to conclude that uncertainty does apply to single particles. 
Does the Heisenberg uncertainty principle apply to the free particle?
Uncertainty principle: for an individual particle?
My question is prompted in part by a "tongue in cheek" video which shows a propeller in a closed box appearing to cause movement. The box is flimsy and the experiment is not meant to be definitive but made me wonder.
 A: Energy and momentum are exactly conserved at the microscopic level. They are not just conserved on the average. This was settled in 1926 by the Bothe-Geiger experiment, which disproved the BKS theory of quantum mechanics, in which conservation of energy was supposed to hold only on the average. (Exact conservation of energy implies exact conservation of momentum, if it holds in all frames of reference.)

Discussions of conservation of momentum frequently use the metaphor of two billiard balls colliding. My impression is that this is not valid at the quantum scale - an illustration of the particles' trajectories should show the outgoing vectors with some uncertainty.

Quantum mechanics doesn't say that all quantities are uncertain. The momentum of a particle can be exactly well defined.
A: If you initially knew the incident momentums, the sum of momentums will be preserved, but its difference almost certainly (it depends in the kind of interaction, but with probability 1) will have some uncertainty, thing that after the collision didn't happen.
