This is just a misunderstanding--- "no motion" in quantum mechanics is a different concept than "no motion" in classical mechanics. At zero temperature, nothing stops. Spherical uncharged black holes don't stop particles at the singularity, they absorb particles and time just ends at the singularity for the infalling matter. The wavefunctions are not made to stop.
You can stop an electron by putting it in the ground state of Hydrogen. This is what it means to be stopped in quantum mechanics.
The reason one can be sure that the uncertainty principle applies to more than what we have seen is that it is impossible to make part of the world classical and part quantum mechanical, as understood by Bohr and Rosenfeld in the early days of quantum mechanics. The fact that the electron has an uncertainty principle means that there would be a contradiction if something else did not, because this would allow you to violate the uncertainty principle for electrons by interacting them with this new thing.
If the world is classical underneath, the classical variables will have very little relation to the position and momentum of classical point particles. This question is annoying, and it does not deserve any more attention than what it has gotten.
The Projection Postulate states that if we have an observable $O$ with discrete spectrum $\{\lambda_i\}$, after a measure in a system resulting in the eigenvalue $\lambda_a$, the initial system $|\psi_i\rangle$ is reducted to the state
$$|\psi_f\rangle=\frac{P_a|\psi_i\rangle}{\|P_a|\psi_i\rangle\|}$$
where $P_a$ is the projection on the eigenspace of the eigenvalue $\lambda_a$. Any repeated measurement of $O$ in this state without time evolution will yield the same result $\lambda_a$, since
$$O|\psi_f\rangle=\frac{OP_a|\psi_i\rangle}{\|P_a|\psi_i\rangle\|}=\frac{\lambda_aP_a|\psi_i\rangle}{\|P_a|\psi_i\rangle\|}=\lambda_a|\psi_f\rangle$$
i.e., $|\psi_f\rangle$ is an eigenvector of $O$ with eigenvalue $\lambda_a$.
In the case of continuous spectrum, there is some complications, since the eigenvectors are not normalizable and hence are not acceptable states. But as the measurement of an exact value of position, for example, cannot be made with arbitrary precision, what is really being measured is the projection over some region of the spectrum, like the dimension of the detector. In that case, say the particle is measured in the region $[a,b]$. By the projection postulate, the final state will be (in position representation, ignoring the normalization):
$$\psi_f(x)=P_{[a,b]}\psi_i(x)=\chi_{[a,b]}\psi_i(x)$$
where $\chi_{[a,b]}$ is the characteristic function of $[a,b]$. That is, the final wavefunction will be the restriction of the initial one to the interval $[a,b]$. Realize that we still have an uncertainty in the momentum as expected, but if the system does not evolve with time, any other measurement of position, will still be restricted to $[a,b]$, since the final wavefunction support is in that region.
Best Answer
For the sake of simplicity, I will answer the question for the bosonic species He(4). Although there are some subtleties for the Fermionic species He(3), due to the presence of total spin-$\frac{1}{2}$, the main message is the same.
The key points are summarized here as follows:
Now we can be quantitative using the harmonic oscillator model. The potential between two atoms is short-ranged repulsive and it becomes attractive for the long-range. Near the potential minimum, the attractive potential can be modeled via the Lennard-Jones potential $-$ $$V(r) = \epsilon_0\left(\frac{d^{12}}{r^{12}}-2\frac{d^6}{r^6}\right),$$ where the parameters $\epsilon_0$ is the trap-depth, i.e., the minimum potential and $d$ is the interatomic separation at the minimum potential.
Since the question involves comparison with other gases, below I put the parameters of He(4) and the closest noble gas neon $$\begin{array}{|l|c|c|} \hline \text{Gas Name} & \text{$\epsilon_0$ [meV]} & \text{$d$ [nm]} \\ \hline \text{He(4)} & 1.03 & 0.265 \\ \hline \text{Neon} & 3.94 & 0.296 \\ \hline \end{array}$$
Now, using the parameters from the above table, we can estimate the zero-point energy in three-dimensions $E_0 = \frac{3}{2}\hbar \omega_0$, assuming an fcc crystal lattice. The oscillation frequency can be estimated as
$$\omega_0 = \sqrt{\frac{4k}{m}},$$ where $$k = \frac{1}{2}\frac{d^2}{dr^2}V(r) = \frac{36\epsilon_0}{d^2}.$$
This expression leads to a $E_0 \approx 7 $ meV for He(4), while the binding energy for atoms is $\approx 1.03$ meV. Therefore the zero-point energy is enough to destroy any crystalline structure of He(4). And this is the reason why He(4) is not found in crystal form, at normal pressure. However, if we compare the binding energy 3.94 meV and the zero-point energy $\approx 4$ meV of neon, we see that the gas can be put into crystal form at relatively small pressure.
To understand the effect of pressure, we look at the following phase diagram of He(4), where we see that the liquid/gas forms continue down to ~0 K, if the pressure remains below 25 atm. The figure distinguishes the two phases He-I and He-II separated by the black line. The superfluid fraction is shown to increase dramatically as the temperature drops.