$\newcommand{\ZFC}{\text{ZFC}}\newcommand{\KM}{\text{KM}}$

The answer must of course be negative, since every model of a theory $T$ is a proper class model of $T$, from its own perspective, but this cannot imply $\text{Con}(T)$ because of the incompleteness theorem.

But the question is actually more problematic than this, since we cannot generally even express that a proper class model is a model of a given theory as a single statement; rather, it is generally a scheme. For example, in the case of the constructible universe, one often hears it said that the constructible universe $L$ is a model of $\ZFC+V=L$, but this is not a single assertion in the language of set theory. Rather, what one means is that one can prove in $\ZFC$ that any given axiom of $\ZFC$ holds also in $L$. Furthermore, because of Tarski's theorem on the non-definability of truth, we aren't really able to formulate the assertion "$L$ is a model of $\ZFC$" as a single assertion in the first-order language of set theory. In this sense, the answer to your question is no.

Meanwhile, however, if you work in a stronger theory, such as Kelley-Morse set theory, then you can recover an affirmative answer. This is because $\KM$ proves the existence of truth predicates for first-order truth relative to any given class. Thus, in $\KM$, if you have a proper class model of a theory $T$, then $\KM$ can build the truth predicate for first-order satisfaction in this model and see that it is consistent, so the answer turns to Yes. The assertion that $L\models\ZFC$, for example, can be formalized as a single assertion in Kelley-Morse set theory, and furthermore, this assertion will imply $\text{Con}(\ZFC)$, essentially by induction on proofs.