Nope.
ZF is consistent with many negations of strong choice. For example, ZF is consistent with Lebesgue measurability of every subset in R. Well-ordering of R is enought to create unmeasurable set.
So, if ZF could prove existence of such a formula, ZF+measurability would prove contradiction, but ZF+neasurability is equiconsistent with ZF and ZF would be inconsistent.
It is very hard to say anything about any well-ordering of R, they are monster constructions...
Here's a sort of related argument (very much not a math expert here): Any well ordering on the real numbers must be non-computable. If there was a computable order, you could establish a 1-1 correspondence between the natural numbers and the reals (since each real would be emitted on the nth step of the algorithm).
Is that remotely right?
From Costanza's original thread (entire text):
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