The answer to the first question is no. First, let $\chi=\forall x>0\\,\exists y\\,(xy=1)$ and $T'=T+\chi$, so that $T'$ is the first-order theory of archimedean ordered fields. Let $\phi(x)$ be a formula defining $\mathbb Z$ in $\mathbb Q$, and let $\psi$ be the sentence “$\phi(x)$ defines a discretely ordered ring”. Since the only DOR embeddable in an archimedean field are the integers, $\phi(x)$ provides an interpretation of true arithmetic in the theory $T'+\psi$. Thus, $T'+\psi$ is not recursively enumerable, and a fortiori $T$ is not recursively enumerable either. (In fact, $T$ is not even arithmetical.)

Let me spell the argument in more detail. For any sentence $\alpha$, let $\alpha^\phi$ be the sentence obtained by relativizing all quantifiers to $\phi(x)$. Then I claim
$$\mathbb Z\models\alpha\iff T\vdash\chi\land\psi\to\alpha^\phi,$$
which implies that $\mathrm{Th}(\mathbb N)$ is recursively reducible to $T$.

Right to left: $\mathbb Q$ is an ordered ring without infinitesimals, hence $\mathbb Q\models\chi\land\psi\to\alpha^\phi$. Also, $\mathbb Q\models\chi\land\psi$ and $\phi(\mathbb Q)=\mathbb Z$, hence $\mathbb Z\models\alpha$.

Left to right: Let $R$ be an ordered ring without infinitesimals, we have to show $R\models\chi\land\psi\to\alpha^\phi$. Assume $R\models\chi\land\psi$. Then $R$ is a field (by $\chi$) and $S=\phi(R)$ is its discretely ordered subring (by $\psi$). Since $R$ has no infinitesimals, it is archimedean, hence so is $S$, which means $S\simeq\mathbb Z$. Thus, $S\models\alpha$, and $R\models\alpha^\phi$.

As for the second question, this is an interesting problem. It may be related to the notorious open problem whether the universal theory of $\mathbb Q$ is decidable (or equivalently, recursively enumerable).