 Solution

Yes: $Q^{+}$ is prime. In any model of its complete diagram, map each rational to the interpretation of its named constant. The diagram ensures this is an embedding, and quantifier elimination for dense linear orders without endpoints makes it elementary. Alternatively every finite tuple of $Q^{+}$ has its complete type isolated by the conjunction equating its coordinates to their named rational constants. Thus it is a countable atomic model, and the earlier prime-model equivalence applies. A prime model may omit many types; primeness is not saturation.