How much of digital sensor noise is thermal?

No. The noise is in the photon count, and it is the square root of the number of photons. So if you get 9 photons, the noise level is 3 (33%); if you get 10000 photons, the noise level is 100 (1%).

Photons are not perfectly distributed over space, but randomly, and even at perfect 0 K, different receiver buckets will catch different amounts of photons each time you try.

There is dark current, as inkista wrote in a comment, and there's quantization noise at the digitizer, and there's gain noise in the analog amplifiers, and there's readout noise (slightly less than 100% of collected charge is extracted).

When people write "scene-limited" or "photon-limited" noise, what they mean is that the photon shot noise is much greater than the combination of all the electronics noise sources. You can never do better than being photon-limited. Luckily, if you read up a bit on energy per photon, you'll see that an SNR of sqrt(number of photons) is a very large number – unless you're imaging distant galaxies or some such :-)

The dominant non-thermal noise sources are photon-shot noise and cosmic rays induced noise, but this only affects images taken by professional astronomers, such as by the Hubble Space telescope. All other noise sources, which then also includes shot noise due to the discrete number of dark current electrons, is ultimately thermal in nature. In ordinary astrophotography, the photon shot noise is always hidden in the thermal noise. E.g. the number of photons from a tenth magnitude star entering a 50 mm aperture lens is about 2500 per second. So, even a 1 second exposure would only have a fluctuation of the photon number of the order of sqrt(2500) = 50, the relative fluctuation is therefore just 2%.

We need to consider a 15th magnitude star (ten thousand times dimmer than the weakest stars visible to the naked eye) to get to a 20% fluctuation due to noise at 1 second exposure. But, of course, in such a picture the star is nowhere to be seen, all you see is read noise and shot noise, the latter is then not induced by the photon number fluctuation, rather by thermal fluctuations in the dark current. By improving the exposure you can reduce these thermal noise effects to make the star visible, but then the photon shot noise will also be reduced.

Suppose that a camera is taking pictures where the noise is dominated by photon shot noise, how we would we know that this is indeed the case? The crucial test would be to take a few dark frame shots. These dark frame shots should then only contain hot and stuck pixels and apart from that almost no noise at all. So, subtracting two dark frame exposures should yield almost totally dark pictures.

A camera cooled to absolute zero would, however, still be affected by read noise and dark current noise besides photon shot noise. This is then caused by quantum fluctuations rather than thermal fluctuations. Any electronic circuit is subject to quantum fluctuations at absolute zero. The simplest example to demonstrate this is an LC circuit, in such a circuit the current and voltage oscillate at an angular frequency of 1/sqrt(LC). When we then describe this system according to quantum mechanical description, we then find that just like any oscillating system, this system has energy levels given by (n+1/2) hbar omega. In the ground state the system then still have an energy due to voltage and current fluctuations of 1/2 hbar omega.