First, treating a proton as an elementary particle is a low energy approximation that can't be used for proton decay. The proton is actually three valence quarks with some virtual gluons and quarks sprinkled about.
Proton as three
point-like particles is an approximation, which is tuned to be treated by mathematical apparatus. I can not see it as three particles u, u, d, but I see it as some
"stable rotating space-time medium" with three mixed poles u, u, d. And you also had wrote that proton do have not only three particles u, u, d, but also “with some virtual gluons and quarks sprinkled about”.
...proton decay...
1. "Proton decay" can occur only in the exited space-time, i.e. in a space-time with changed geometry, or, which is the same, - in the strong fields.
2. As a result, it can not be thought as if it decays per quarks. Three-polar construction of proton under its decay creates two two-polar constructions: x-boson and positron. The decay process is accompanied by ejection of a couple of gamma-quanta. At this figure we can see a proton as three-polar construction (u,u,d):
After entering into a strong field (or in Kaluza Klein space-time) and after proton decay we can see two two-polar constructions:
positron, having two magnetic poles;
and x-boson, also having two poles: magnetic NS; or electromagnetic N+S-, or N-S+, or N~S~, or mixed electro-magneto-weak. These poles are not already the QCD’s quarks, because x-boson can only live in this unusual geometry.
Second, the X boson has a charge of 4/3. The charge of your proton decay equation isn't adding up.
My x-boson do not consist of two up quarks. It has some properties of X and Y bosons; and it has some properties of “Kaluza-Klein Z’, W’ bosons”.
Third, the X boson doesn't need a "magnetic hole" to cause proton decay. Two up quarks can merge into a virtual X boson, which can then decay into a positron and an anti down quark.
http://en.wikipedia.org/wiki/X_boson
My x-boson is not a virtual pair of up quarks, but a real two-polar construction, which can exist only in the strong field (only in Kaluza-Klein space-time). In order to make it free, one must spent energy, which is more then binding energy of x-boson in magnetic hole. If such energy (E) is added, then x-boson goes out and decays like this:
x-boson + E --> p + e.
The needed energy E ~ m[sub]pr[/sub]c[sup]2[/sup]/3.
The low observed rate of proton decay shows that the mass of the X boson must be around 10^15GeV or higher.
Proton decay can occur only in already created strong field. The mass of my x-boson is about (2/3) of proton mass. But the minimal possible magnetic hole, which is able to ruin protons, must have about 340 of such x-bosons. That mass corresponds to about 0.34 TeV. Total binding energy of such hole is about 0.17 TeV. In order to create such hole, it is necessary to collide two protons with energy about 0.25 TeV per proton.
Virtual particles of that type would cause proton decay, and the observed low rate of proton decay shows that such particles must be extremely massive, far beyond anything the LHC can ever do.
Observed low rate of proton decay is close to zero because there were no p-p collisions of corresponding energy yet. Wait a little, soon, the process of extremely high rate of proton decay will be switched on and the Solar system will look after as the remnants of SN 1987A.