On the stability of deuterium
No, deuterium is completely stable. I found the answer at Hyperphysics, and it has to do with the mass energies of the products and reactants of this hypothetical reaction.

The decay of deuterium would be 
$$\ce{D -&gt; P + N + e + \bar{\nu}}_e$$
where $\ce{D}$ is deuterium, $\ce{P}$ is a proton, $\ce{N}$ is a neutron, $\ce{e}$ is an electron and $\ce{\bar{\nu}}_e$ is an electron antineutrino. The combined mass energies total to $1877.05\,\mathrm{MeV}$ (mega electron volts). But the mass of $\ce{D}$ is $1875.6\,\mathrm{MeV}$, so the decay cannot happen because conservation of mass/energy would be broken. There's no way to get around this, not even imbuing the initial deuterium atom with extra energy of some other sort. Deuterium will never decay unless, for some reason, the proton turns out to be unstable.

If, for some reason, deuterium did decay, it could go to $\ce{^2He}$ … which would then decay back to deuterium.