The short answer is it depends on just what you mean. The long answer:
When you're looking at a nonstandard model of arithmetic, there's two perspectives from which you might make a definition: externally or internally. From the external perspective, the only real formulae are the finite, well-founded ones, those with a Gödel code in $\mathbb N$. So it doesn't make sense to ask whether some nonstandard $\varphi$ is true or not, since it's not really a formula in the real world.
From the internal perspective, you're looking at what is expressible in the model. A nonstandard model $M$ of PA doesn't know that $\varphi$ is really a weird infinite "formula"; from $M$'s perspective $\varphi$ satisfies the definition of being a (Gödel code of) formula in the language of arithmetic. So does $M$ think $\varphi$ is true?
Since we're thinking from the internal perspective, we must confine ourselves to what $M$ can express. And this is where we run into a difficulty. It's a theorem of, as I recall, Mostowski that if $k$ is a standard integer then the truth predicate restricted to formulae of depth $\le k$ is definable. (The notion of depth that's usually most useful is the number of alternating unbounded quantifier blocks.) For standard formulae $\varphi$ we can make sense of truth internally. (Though not uniformly across all formulae, only across a bounded class.) But if $\varphi$ is nonstandard it won't have standard depth, so we can't use any of those.
Indeed, it's a corollary of Tarski's theorem on the undefinability of truth that you cannot have a definable truth predicate for all formulae of depth $\le e$ for some nonstandard $e$. The reason is, if it were definable, then it'd be definable by some formula $\theta(\varphi,x)$, but $\theta$ is a real formula and so $M$ thinks it has depth $\le e$, and then you can diagonalize. In short, if we just look at what $M$ alone can see then it doesn't make sense to ask whether $\varphi$ is true or not.
But we can expand a bit. We can look instead at $M$ adjoined with a class $T \subseteq M$ so that $T$ satisfies the Tarskian definition of a satisfaction class. Call this $T$ a full satisfaction class, since it measures the truth of all of what $M$ thinks are formulae. This starts to get heavily into the model theory of arithmetic, so let me just state a few highlights.
If there is a full satisfaction class $T$ for $M$, then $M$ is recursively saturated. In particular, not every $M$ admits a full satisfaction class.
If $M$ is countable and recursively saturated, then there are continuum many different full satisfaction classes on $M$. So depending on which $T$ you pick $\varphi$ might be either true or false.
There are uncountable, recursively saturated $M$ which don't admit any full satisfaction class.
The definition of recursive saturation, as well as proofs of all these can be found in Richard Kaye's excellent Models of Peano Arithmetic, in chapter 15. In general, Kaye's book is the go-to source for questions about nonstandard models of arithmetic.
