I do not have the reputation to comment, but I'll delete this after you've read it.
There is an idea, mostly from virtual knot theory, called parity (I recommend this reference: arxiv:1211.0403, and I won't bother copying the definition). Fiedler and Stoimenow's "New Knot and Link Invariants" paper from the Knots in Hellas proceedings' book gives you an example of a parity for classical braid (they call it type), and some other exist.
I think that a parity projection proof might work to show that in fact no such shortcut can exist. You just need to define the appropriate parity.
I know about parity, but can't see right now how it would lead to a proof- anything more you can say would be great!
The standard idea of a parity-projection proof is that you find a parity which singles out crossings that do not satisfy the property you want and then an arbitrary sequence of Reidemeister moves between two even diagrams. The Reidemeister moves which create odd diagrams are ignored, usually by turning them into the corresponding virtual move. Then, by the axioms of parity, the resulting sequence of diagrams, which each differ by a single extended Reidemeister mover all satisfy the desired property.
In the case that you want, using virtual crossings is a bad idea because an isolated crossing would be odd, and that would restrict the next moves. (Think of doing a RM2, then a RM on a arc between the two new crossings, then a RM3. If the RM1 is changed to a virtual RM1, the RM3 is no longer allowed without going to unwelded string links, and classical braids do not embed inside those.)
So, my idea would be to use a projection that smooths the crossing according to the orientation of the string link. We still have to define an appropriate parity.
Now, this is the part where I am not sure the math actually works, since this construction does not respect the parity axioms, one still has to check that this projection isn't too destructive. I will call this the monotonicity parity, denoted $m(c)$. Let $c$ be a crossing is a string link $L$, and $L_c^+$, $L_c^-$ be the components of $L$ smoothed at $c$ (again, with orientation) that contain the arcs that were smoothed. The, $m(c)=0$ if and only if both $L_c\pm$ are long components.
It is quite obvious that this parity is invariant under Reidemeister moves away from the crossing smoothed and planar isotopies, and not too complicated to see that two crossings that can cancel in a RM2 have the same parity and an isolated crossing is odd. The hard part is RM3. The parities of the crossings involved in one is invariant under the move, but what about the requirement that the number of odd crossings in a RM3-like configuration not be odd? It isn't satisfied, precisely by the example sequence I gave above, given that you had a braid after the RM2. This is were everything goes wrong in the general case, but possibly not in the case you care about, since the odd crossings created by moves from a braid diagram are unknots which aren't linked with the rest of the string link.
Possibly, these unknots can be erased, and any future RM that involved them ignored, while still obtaining the same final braid diagram.