The production of neutrons by D-D side reactions also produces tritium in further subsequent side reactions- well may be. In the Polywell at least, the tritium and proton produced from D-D fusion are above the potential well energy (presumably, unless you are going to extreme KE injection energies). They leave the system. As pointed out multiple times in various threads, the Polywell is not an ignition machine - the fusion products do not contribute significantly to further fusion or plasma heating. They have to be collected externally, processed and reintroduced as new injected fuel ions.
The same would presumably hold for neutrons. Despite higher neutron absorption by eg: 10B, there is very little Boron in the reaction volume. The plasma may have a density of perhaps one thousands of an atmosphere (10^22 particles/ M^3). That this relatively low density of particles can lead to significant fusion is helped by the density squared scaling, but also critically by a confinement time represented by the number of passes across the machine(along with the speed). At least 10 thousand passes within a few meter diameter machine is needed at 10^22/ M^3 densities. A neutron only has one pass, or rather probably only 1/2 pass as most fusion produced neutrons are released in the central core of a working Polywell reactor. This is a confinement time/ distance difference of at least 20,000. The neutron captures by 10Boron or any other isotope in the reaction space is extremely tiny in relation to the parent fusion rate. This would hold for any fusion product reaction , even ignition plasma where the fusion produced ions may still be magnetically and/ or electrostatically confined at times >> greater than absolutely unconfined neutrons.
The 1/2 cat D-D reaction has been described by Bussard and others(?). Here there is a 10B blanket surrounding the reactor which captures the escaping neutrons, with heat and and tritium production contributing to the final desired energy balance. I don't know the thickness of this solid blanket, but an exercise with guessed numbers may illustrate the possible contribution of neutron absorption inside the plasma. Assume the 10B solid blanket is several inches thick. A gas of the same intercepting mass/ number of atoms would have to be compressed about 700 fold (700 atmospheres) to have the same density. As the plasma is probably about 0.001 atm (10^22particles/ M^3 at STP) results in a final difference of ~700,000. For the plasma to absorb as many neutrons as the solid exterior blanket, the plasma reaction volume would need a radius of perhaps 1,000,000 inches or about 30 Kilometers.
I have a nuclear fallout calculator wheel buried in a box somewhere, but I recall that perhaps several feet of dirt will protect against fallout radiation- mostly gamma rays and neutrons, but hundreds of yards or more of air is required. And the typical plasmas are at a thousandth of an atmosphere, or in a Tokamack example perhaps a millionth....
Keep in mind that we are talking density, not pressure- so don't get too confused by my jumping between the units. One atmosphere of gas at STP has density of about 10^25 particles per cubic meter. Heating to a higher temperature gas, like a plasma, may result in a much higher pressure in a confined space (PV=nRT), but the comparative densities are the same. This may be misleading if relationships are not accounted for- such as MIT's recent claim that their Tokamak reached over one atmospheric pressures. At their temperatures, this was still the mundane ~ 10^20 particles per cubic meter (still better than 10^19 particles per cubic meter of some Tokamak results)
To error is human... and I'm very human.