= Solution
For <cut types over the named rational order>, choose an irrational real cut, for example $r=\sqrt2$, and form
$$
p_r(x)=\{c_q<x:q\in\mathbb Q,\ q<r\}\cup\{x<c_q:q\in\mathbb Q,\ q>r\}.
$$
Every finite <subset> is realized by a rational between its finite lower and <upper bounds>, so compactness makes it a <consistent first-order theory>[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> <first-order model>[model] of $T$ realizing it. An <isomorphism of structures>[isomorphism] in $L_{\mathbb Q}$ fixes all named rationals and preserves realization of the cut, so it cannot be isomorphic to $Q^+$. Thus \b[$T$ is not countably categorical]. Omission by itself would need this additional realization-and-countable-model argument to establish noncategoricity.
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