As Joel said, most of the consistent extensions $T$ of ZFC are not in the minimal transitive model $M$ of ZFC and therefore don't have models in $M$. It seems wortth noting that this is the only reason the answer to the question is negative. If a consistent extension $T$ of ZFC is an element of $M$, then it has a model in $M$. The proof of this just consists of the known facts that $M$ satisfies "$T$ is consistent" (i.e., consistency is absolute for $M$) and Gödel's completeness theorem is provable in ZFC and therefore holds in $M$ (and additional absoluteness: Whatever $M$ considers to be a model of $T$ really is a model of $T$).

As Joel said, most of the consistent extensions $T$ of ZFC are not in the minimal transitive model $M$ of ZFC and therefore don't have models in $M$. It seems worth noting that this is the only reason the answer to the question is negative. If a consistent extension $T$ of ZFC is an element of $M$, then it has a model in $M$. The proof of this just consists of the known facts that $M$ satisfies "$T$ is consistent" (i.e., consistency is absolute for $M$) and Gödel's completeness theorem is provable in ZFC and therefore holds in $M$ (and additional absoluteness: Whatever $M$ considers to be a model of $T$ really is a model of $T$).