In principle
Why shouldn't a BEC be stable? A BEC in free space can be described as a solution to the Gross-Pitaevski equation, hence it is an eigenstate of the Hamiltonian. It is then a stationary state, i.e. it keeps being an eigenstate of the Hamiltonian for all time.
This assumed you are not perturbed by any external influence, hence that no decoherence comes in.
A canonical way of solving this is via the "mean-field" approach, which sees the introduction of the creation and annihilation operators $a$ and $a^{\dagger}$.
In reality
A BEC is a metastable state of matter, since the "stable" state is a solid. Two atoms in a BEC can form a molecule, releasing their binding energy as kinetic energy of a third atom - these are called 3-body collisions, and scale as $n^2$, where $n$ is the atomic density. The molecule is basically the beginning of the formation of the stable molecular solid, while the third atom has usually enough energy to escape whatever trap the atomic cloud was in, heating other atoms in the process.
So the BEC has a lifetime $\propto 1/n^2$, which why it only exists as a dilute gas.
Also, you can't create a BEC in free space, you always need to trap it - its nonzero temperature means that it would always spread around. The external trap modifies the system such that the BEC will be in the ground state of this external potential, however you are now coupled to an external factor and subject to its potential instabilites.
Finally, a BEC only exists in a vacuum since you want to minimise interactions with non-BEC atoms (such as air), both to avoid heating and to avoid decoherence. The fact that there is no such thing as a perfect vacuum, means that experimentally your BEC will always have a finite lifetime, which can range from tens of seconds to tens of minutes.
