A seemingly problematic aspect of the proposed mechanism is that it allegedly requires two hot deuterons. (By contrast, U-235 fission requires just one neutron.)
Why is that so problematic? If $n$ is the number of 20keV particles (i.e. hot deuterons, or K-shell holes, or some superposition of them), then we expect something like:
$$dn/dt = An^2 - Bn$$
where the coefficient $A>0$ describes fusion and $B>0$ describes cooling into lower-energy modes.
This describes a very badly behaved chain reaction. This differential equation supports explosive growth of the number of hot deuterons, and it supports very easily fizzling out entirely. I don't see how it could support a reaction that goes on for 50 hours, which is the alleged observation in cold-fusion experiments. (You can object that it moves from hot-spot to hot-spot within the electrode, but even so, I find it implausible. I wouldn't expect local hot-spots; I would expect a fast-growing hot region that would fuse almost every deuteron in the entire electrode within a second! Unless I'm mistaken...)
A U-235 reaction, where there is no $n^2$ term, is relatively easy to stabilize. In theory, all you need is a negative temperature coefficient. But even a negative temperature coefficient would fail to stabilize this kind of quadratic reaction rate (if I'm not mistaken).
I won't say this disproves the mechanism, but I would say that this is something that warrants explanation and discussion.
I also wonder whether you really need the energy of two K-shell holes, not just one, to overcome the Coulomb barrier. If one hole is enough, then the problem above does not apply. Moreover, you wouldn't really need to worry about hot-deuteron-lifetime (the other major potential problem with the mechanism), because maybe the hot deuteron doesn't travel around the lattice at all. Maybe there is a K-shell hole and two deuterons all together in the lattice, and the hole's energy simply shoves one deuteron into the other. (The hole's energy becomes Coulomb potential energy, not kinetic energy.)
Again, I don't know if 20keV is enough energy. But if it were, that would make the story much more plausible in my opinion. :-D
Your intuition about the charge repulsion and strong force acting on Protons more is less important that you think. The strong nuclear force is a few orders of magnitude greater than electromagnetism so coulomb repulsion just doesn't contribute much.
What matters most is the nuclear binding energy to separate a proton from the nucleus. If the resulting system is below the proton separation energy it's possible for the proton to tunnel out. See Proton emission and Proton drip line for more information about this. It does happen but remember Neutron emission is also rare. $\beta^+$ and $\beta^-$ and Alpha emission are much more common.
Best Answer
According to your question and the discussion in the comments there seems to be some confusion. Let's first try to answer your original question (Coulomb barrier for a D-D reaction):
The Coulomb barrier is usually defined as the point where the strong force overcomes the repulsive Coulomb force of two positively charged nuclei. A proper estimation is that the nuclei need to barely touch, thus have a distance of $$R = R_1 + R_2,$$ where $R_1$ and $R_2$ are the radii of the nuclei. As you pointed out, a useful approximation for a nucleus' radius is $$R=r_0 A^{1/3},$$ with $A$ the mass number and $r_0\approx1.3\cdot10^{-15}\,\mathrm{m}$. The distance at which the strong force starts to set in can thus be written as $$R\approx r_0\left(A_1^{1/3} + A_2^{1/3}\right).$$ Let's now plug that into the Coulomb potential: \begin{eqnarray} V_{Coulomb}&\approx&\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r_0\left(A_1^{1/3} + A_2^{1/3}\right)}\\ &=&\frac{e^2}{4\pi\epsilon_0}\frac{Z_1Z_2}{r_0\left(A_1^{1/3} + A_2^{1/3}\right)}, \end{eqnarray} with $q_1$, $q_2$ the charge of the nuclei and $Z_1$ and $Z_2$ their charge numbers.
Inserting the numbers for a D-D reaction, this results in $$V_{Coulomb}\approx440\,\mathrm{keV},$$ which corresponds to the value you got.
This corresponds to the potential wall we need to overcome to let the two Deuterium nuclei fuse. Luckily, though, there are two effects leading to a on average lower temperature at which fusion occurs:
The fusion rate coefficient or reactivity, $\left<\sigma_{fus}v\right>$, can therefore be calculated as the convolution of a Maxwellian distribution $f_M$ and the fusion cross section $\sigma_{fus}$ (which we know from extensive experiments). So to get the average temperature at which D-D fusion has the highest probability, we need to have a look at the fusion reactions:
They both occur with roughly the same probability. To calculate their actual reactivity, we look up data from experiments and might get something like what is shown in the following plot.
As you can see, the rate coefficient of the D-D reaction is basically increasing with energy (temperature), it will asymptotically approach a very broad maximum at temperatures which are larger by an order of magnitude (sorry for not showing but I couldn't find the necessary data). For comparison, the D+T reaction is also shown and you can see why this is favored in the lab: its reactivity is larger by two orders of magnitude almost over the full range shown here.
To summarize, you don't need to actually climb up all the Coulomb wall to let the particles fuse, finite probability for fusion does occur at much lower temperatures.