A 'transfer principle' between models of $\mathbf{ZF+AR}$ and $\mathbf{ZF}$ For convenience, I shall not consider the axiom of regularity as being an axiom of $\mathbf{ZF}$ and refer instead to $\mathbf{ZF+AR}$ when including regularity.
Suppose $\mathfrak{M}\models\mathbf{ZF}$ then my question is if for every sentence $\sigma$ not belonging to the deductive closure of $\mathbf{ZF+AR}$ there exists a $\mathfrak{M}_{WF}\models\mathbf{ZF+AR}$ such that
\begin{equation}
\mathfrak{M}_{WF}\models\sigma \quad\Longrightarrow\quad \mathfrak{M}\models\sigma
\end{equation}
or not?
 A: The question as written allows us to pick a different $\mathfrak{M}_{WF}$ for each sentence $\sigma$ and does not require any connection between $\mathfrak{M}$ and $\mathfrak{M}_{WF}$. It seems very unlikely that this is what you had in mind. But as written, the answer is yes:
Suppose $\mathfrak{M}\models \mathbf{ZF}$. Fix a sentence $\sigma$ which is not provable from $\mathbf{ZF+AR}$. Then $\mathbf{ZF+AR+\lnot\sigma}$ is consistent, so it has a model, which we'll call $\mathfrak{M}_{WF}$. Then the implication $$\mathfrak{M}_{WF}\models \sigma \implies \mathfrak{M}\models \sigma$$ holds because the premise is false: $\mathfrak{M}_{WF} \models \lnot \sigma$. 
Note that the chosen model $\mathfrak{M}_{WF}$ depended entirely on $\sigma$ and not at all on $\mathfrak{M}$. Again, I can't believe this is what you had in mind. 

So I'll take a guess that you got the quantifier order backward. Maybe you wanted to ask: 

Suppose $\mathfrak{M}\models \mathbf{ZF}$. Does there exist a model $\mathfrak{M}_{WF}\models \mathbf{ZF+AR}$ such that for every sentence $\sigma$ not provable from $\mathbf{ZF+AR}$, we have $\mathfrak{M}_{WF}\models \sigma\implies \mathfrak{M}\models \sigma$?

Of course if $\mathfrak{M}\models \mathbf{AR}$, then we can take $\mathfrak{M}_{WF} = \mathfrak{M}$. But if not, the answer is no. Fix $\mathfrak{M}\models \mathbf{ZF+\lnot AR}$, and assume for contradiction that $\mathfrak{M}_{WF}$ exists.
Let $\varphi$ be any sentence which is independent from $\mathbf{ZF+AR}$. For example, we could take $\varphi$ to be $\text{Con}(\mathbf{ZF+AR})$ or its negation. (Note that the existence of $\mathfrak{M}$ implies that $\mathbf{ZF}$ is consistent, from which it follows that $\mathbf{ZF+AR}$ is consistent, and hence Gödel's Second Incompleteness Theorem tells us that neither $\text{Con}(\mathbf{ZF+AR})$ nor its negation is provable from $\mathbf{ZF+AR}$). 
Without loss of generality (replacing $\varphi$ with $\lnot \varphi$ if necessary), we can assume that $\mathfrak{M}_{WF}\models \varphi$. Now let $\sigma = \mathbf{AR}\land\varphi$. Note that also $\sigma$ is not provable from $\mathbf{ZF+AR}$, since if it were, $\varphi$ would be too. But $\mathfrak{M}_{WF}\models \sigma$, so by transfer $\mathfrak{M}\models \sigma$. This contradicts the fact that $\mathfrak{M}\models \lnot \mathbf{AR}$. 
