While I do not have an actual answer, the following observation may be useful:
Assuming Global Choice the answer is trivially yes. You can preform Scott's trick on order type equivalence classes, and choose the representative accordingly.
Assuming the axiom of choice, it is always possible to assign a set collection of order types. This set can be made as large as you want, however we are not guaranteed to have a class of representatives.
The ordinals are definable in ZF (without the axiom of choice at all) due to von Neumann's axioms of foundation and replacement schema (both attributed to Frankael and Skolem). This allows the unique definition of well-order types using transitive $\in$ ordered sets.
On the other hand, there is no way to define general partial orders in ZF or even ZFC in a way which would result in a definable representative to each class.
Much like Jech's proof that without the axiom of choice it is possible to have a model of ZF which has a set of cardinalities (defined as per Scott's trick) which has no definable choice of representatives, I would believe that such proof would be possible to achieve via forcing in ZFC.
It is immediate that such claim is likely to be independent of ZFC, since V=L implies Global Choice, which in turn implies that there is a canonical representative. However if it is possible to construct a model in which there is a class of order types without definable representatives (and I very much believe that it is possible) then it is independent of ZFC.