With regard to the deeper motivation behind this question: I think there's some confusions around exactly how to interpret nonstandard models. I'm going to focus on the very narrow question, "Can models disagree about the computability of certain definable sets?"
The first step is realizing that this is a bit vague. There's at least three ways I can see to interpret it:
Interpretation 1 (Internal). You give me a formula in the language of arithmetic - viewed as a definition of a set - $\varphi(x)$. Each model $M$ of $PA$ yields a set $\varphi^M$. Now, $\varphi^M$ is a subset of $M$, not $\mathbb{N}$, but $M$ can't tell the difference. The sentence $\psi\equiv$"$\{x: \varphi(x)\}$ is decidable(=computable)" is expressible in the language of arithmetic, as $$\psi\equiv\exists e_0, e_1\forall x[(\varphi(x)\iff \exists sT(e_0, x, s))\wedge (\neg\varphi(x)\iff\exists sT(e_1, x, s))],$$ where $T$ is Kleene's $T$ predicate: $T(e, x, s)$ means "the $e$th Turing machine halts on input $x$ in $s$ steps."
Okay, fix $\varphi$. We can now ask:
Is there a model $M$ which satisfies $\psi$? $\neg \psi$?
Interpretation 2 (External). You give me a formula in the language of arithmetic - viewed as a definition of a set - $\varphi(x)$. Each model $M$ of $PA$ yields a set $\varphi^M$. Now, $\varphi^M$ is a subset of $M$, but the standard model $\mathbb{N}$ injects into $M$ in a canonical way; call this map "$i_M$." So we can talk about the standard natural numbers which ($M$ thinks) the statement $\varphi$ is true of; that is, the set $$X_M=i_M^{-1}(\varphi^M).$$ This set $X_M$ is a genuine set of natural numbers, and we can ask whether $X_M$ is: computable, infinite, coinfinite, of asymptotic density 1, etc. For instance, in the Goodstein example, $X_M$ is the set of standard natural numbers which $M$ thinks have halting Goodstein sequences.
This "cutting off" process might seem unnatural, but it's actually really mathematically interesting: it's what standard systems, for instance, are all about.
Okay, fix $\varphi$. We can now ask:
Is there a model $M$ such that $X_M$ is computable? Non-computable?
Interpretation 3 (Mix-and-match). As in interpretation 1, we look at the whole set $\varphi^M$. However, we now look at standard indices for computable functions, interpreted in $M$. Specifically, say $\varphi^M$ is mix-and-match-computable in $M$ if there is there some standard natural number $n$ such that $M$ thinks $\Phi_n$ is a total computable function such that $\Phi_n$ is the characteristic function of $\varphi^M$.
Okay fix $\varphi$. We can now ask:
Is there a model $M$ such that $\varphi^M$ is mix-and-match computable in $M$? Non-mix-and-match computable in $M$?
Despite the heterogeneous nature of this question, I suspect it too has its place in your intuition: basically, this question is asking, "Is there a genuine algorithm which, consistently, solves this problem?"
Now, what about your actual question? Well, we can exhibit formulas $\varphi$ such that there are models exhibiting divergent behavior in each of the three interpretations. This is actually pretty easy to do: we just shoehorn in some known undecidability. For instance, consider the formula $$\varphi(x)\equiv [\neg Con(PA) \wedge x\not=x]\vee[Con(PA) \wedge \exists s(T(x, x, s)].$$ In a model of $PA$ which thinks $PA$ is inconsistent, $\varphi$ just denotes the empty set, but in any model of $PA$ which thinks $PA$ is consistent - such as the standard one - $\varphi$ denotes that model's halting problem, which that model will think is incomputable. This might seem a bit trivial - it certainly does to me - but there's not an obvious way I can see to "carve out" such trivial solutions. (Maybe someone here can think of one!)
Best Answer
Assume $\mathsf{ZFC}$ is $\Sigma^0_1$-sound.
For each sentence $\varphi$ in the language of set theory, consider the machine $T_\varphi$ which - regardless of input - searches for a $\mathsf{ZFC}$-proof or disproof of $\varphi$ and halts iff it finds one. Then $T_\varphi$ halts iff $\varphi$ is $\mathsf{ZFC}$-decidable, and $\mathsf{ZFC}$ proves this.
Since $\mathsf{ZFC}$ is $\Sigma^0_1$-complete, if $T_\varphi$ halts (on input $0$, say) then $\mathsf{ZFC}$ proves that $T_\varphi$ halts. What if $T_\varphi$ doesn't halt? Well, then $\mathsf{ZFC}$ can't prove that $T_\varphi$ does halt by the soundness assumption above, and by the last four words of the previous paragraph + Godel's second incompleteness theorem $\mathsf{ZFC}$ can't prove that $T_\varphi$ doesn't halt either (otherwise $\mathsf{ZFC}$ would prove that $\mathsf{ZFC}$ is consistent).
So the halting status of $T_\varphi$ is $\mathsf{ZFC}$-decidable iff $\mathsf{ZFC}$ decides $\varphi$. But it's a standard exercise that the set of $\mathsf{ZFC}$-decidable sentences isn't computable, since otherwise we could build a computable consistent complete extension of $\mathsf{ZFC}$ contradicting the first incompleteness theorem.