 Solution

For cut types over the named rational order, choose an irrational real cut, for example $r=\sqrt{2}$, and formEvery finite subset is realized by a rational between its finite lower and upper bounds, so compactness makes it a consistent partial type; quantifier elimination for dense linear orders without endpoints makes these comparisons determine a complete type in the language naming all rationals. No rational realizes the cut, so $Q^{+}$ omits it.

Add a new constant realizing this type and apply compactness, followed by downward Löwenheim-Skolem in the countable expanded language. This gives a countable model of $T$ realizing it. An isomorphism in $L_{\mathbb{Q}}$ fixes all named rationals and preserves realization of the cut, so it cannot be isomorphic to $Q^{+}$. Thus $T$ is not countably categorical. Omission by itself would need this additional realization-and-countable-model argument to establish noncategoricity.