The answer is negative.
Consider the model $M=\langle\mathbb N\cup\{\infty\},0,S,+,\cdot\rangle$, where we put $S(\infty)=\infty$, $\infty+x=x+\infty=\infty$ for all $x\in M$, $\infty\cdot0=0\cdot\infty=0$, and $\infty\cdot x=x\cdot\infty=\infty$ for $x\ne0$. It is easy to check that $M\models Q$. In fact, $M$ satisfies the axioms of commutative semirings, hence any term is in $M$ equal to a polynomial with nonnegative integer coefficients, and these polynomials can be manipulated in the expected way.
Lemma: The set of sentences $\phi$ of the form $\exists y_1,\dots,y_m\,f(\vec y)=g(\vec y)$ valid in $M$ is decidable.
Proof: By the remark above, we can write $f,g$ as polynomials in $\mathbb N[\vec y]$. Notice that if $h\in\mathbb N[\vec y]$ is non-constant, we have $h(\vec\infty)=\infty$.
Case 1: If $f$ and $g$ are nonconstant, then $M\models\phi$, as witnessed by $\vec y=\vec\infty$.
Case 2: Let (wlog) $g$ be constant, say $g=c\in\mathbb N$. I claim that if $f(\vec a)=c$ for some $\vec a\in M$, then also $f(\vec b)=c$, where $b_i=\min\{a_i,c\}$. Indeed, if $h(\vec y)=\prod_{i\in I}y_i$ is a monomial that appears in $f$ with a nonzero coefficient, and $a_i>c$ for some $i\in I$, then $a_j=0$ for some $j\in I$, lest $f(\vec a)\ge h(\vec a)>c$. Thus, $h(\vec a)=h(\vec b)=0$. Consequently, $M\models\phi$ iff there are $a_1,\dots,a_m\in\{0,\dots,c\}$ such that $f(\vec a)=c$, and this can be algorithmically checked.$\qquad\Box$
Now, let $T$ be the theory axiomatized by $Q$, the axioms of commutative semirings, and $\{\phi:M\models\phi\}\cup\{\neg\phi:M\nvDash\phi\}$ for Diophantine sentences $\phi$. Then $T$ is consistent (being true in $M$), and recursively axiomatized (by the lemma), but for every Diophantine formula $\phi(x)$ and $n\in\mathbb N$, the sentence $\phi(S^n(0))$ is decidable in $T$.
Let me remark that while the answer above exploits the weakness of $Q$ which allows for quite pathological models, reasonable stronger base theories can still make a trouble. In particular, it is a long-standing open problem whether the universal fragment of the theory of quantifier-free induction ($\mathit{IOpen}$) is decidable; if (contrary to expectation) it is, then one can construct a counterexample $T$ as above with $T\supseteq\mathit{IOpen}$.
On the other hand, the answer is positive for theories $T$ extending $I\Delta_0+\mathit{EXP}$, as this theory proves the MRDP theorem.