As you correctly say, QCD flux tubes may snap because (light) quark-antiquark pairs may be created. In string theory, open strings must satisfy the condition that the end points are attached to a D-brane - which may have any dimension as you equally correctly wrote. The endpoints of open strings are points within D-branes that may be identified with quarks or antiquarks in QCD in this (more than) analogy.
The idea of creating a D0-brane-anti-D0-brane pair is creative but it doesn't work because D0-branes - and other D-branes - are too heavy and the string simply can't have enough energy to create these new heavy objects (unless the string carries so much vibrational energy on it that the perturbative expansions are hopelessly broken).
The D0-brane mass goes like $m_s/g_s$ which is the same mass as $1/g_s$ times the string length of a string. One would need to use a very long piece of string to get the required energy for the creation of the new D0-brane and its antiparticle - and this won't happen. This is true not just for D0-branes but all D-branes - all their tensions are proportional to $1/g_s$.
Alternatively, a D0-brane carries the same mass/energy as a highly excited string mode whose excitation level is $1/g_s$ - well, in fact, it has to be $N\approx 1/g_s^2$ because it's the squared mass that $N$ is proportional to. For such super excited modes, the perturbative approximation of string theory doesn't work. Indeed, the fact that non-perturbative objects such as D0-branes may emerge out of the energy of strings is a way to prove that the perturbative expansions fail.
Non-perturbatively, for large enough values of $g_s$, anything can happen. The perturbative expansions fail, D-branes become as easy-to-be-created as the (no longer) fundamental strings, and there are lots of open strings attached to cheap D0-branes and other D-branes flying in a region with not so high energy.
