If not, are there any conditions under which there must be a model under ZFC theory? Alternatively, is there any set of axioms for which this does hold true?
If so, can we drop some of the axioms and still have this power? Are there other proposed foundations of mathematics with this ability?
If the answer to this question is unknown, is it widely accepted (or rejected) by the mathematical community? If either, why?
Gödel's Completeness Theorem states that any consistent theory has a model. This is provable in ZFC (in fact, in much much less than ZFC), so certainly we have:
However, note the "ZFC proves" part. If $T$ is a consistent theory, but ZFC can't prove that, then ZFC can't prove that $T$ has a model. For instance, we usually assume that ZFC is consistent. But, by Gödel's Incompleteness Theorem, ZFC can't prove that ZFC is consistent.
Since ZFC proves Completeness, though, this is the only obstacle.
You may be interested in the question, "What sort of axiom systems prove Completeness?" This has been studied in a few different ways; for example, if we are interested in theories in a countable language, then the very very weak theory $WKL_0$ is already strong enough to prove the Completeness theorem. We really only need set theory to handle cases where the language is "large," and even then we don't need very much.