Ok, I'm pretty sure the answer is no to the basic question (still not sure about the generalization).  We build a set $A$ to satisfy the following conditions.

$P_e: A \neq \phi_i(0^{n})$
$R_{i,j}: X = \phi_i(0^\omega) \land (\forall i)\left(X(i) = A \models \psi_j(i)\right) \implies (\exists n)(X \leq_T 0^{n})$

Where $\phi_i$ is a Turing reduction and $\psi_j$ is an arithmetic formula with one free variable.

We build $A$ via sequence of perfect trees $T_n$ with $T_{n+1}$ a subtree of $T_n$ as in the construction of a minimal arithmetic degree to meet the above requirements with $\lvert T_n(0) \rvert \geq n$.  We ensure that each $T_n$ is arithmetic and is uniformly computable from $0^{\omega + 1}$ and set $T_{-1} = 2^{< \omega}$

To meet $P_e$ we simply insist that $T_e(0)$ disagrees with $\phi_i(0^{n})$.  Given an index for $T_{e - 1}$ we can find an index for a subtree $\hat{T}_e$ with this property and with $\lvert T_n(0) \rvert \geq e$ computably in $0^{\omega}$.

To meet $R_{i,j}$ with $e = \langle i,j \rangle$ now ask if there is a string $\tau$ on $\hat{T}_e$ and an $x$ such that $\tau \Vdash \psi_j(x) = k \land \phi_i(0^{n};x) = 1 -x$ (here we are using local forcing on $\hat{T}_e$).  If so we let $T_e$ be the subtree of $\hat{T}_e$ above $\tau$ and otherwise let it be $\hat{T}_e$.  Note that this whole construction can be carried out computably in $0^{\omega +1}$

It's obvious that the requirements $P_e$ are satisfied and we've constructed $A \leq_T 0^{\omega+1}$ so $A$ clearly satisfies $0 <_a A \leq_A 0^{\omega}$.  If $R_{i,j}$ is also satisfied then we are done since any $\widehat{A} \leq_a A$ with $\widehat{A} \leq_T 0^{\omega}$ would necessarily be arithmetic and therefore $A$ couldn't be arithmetic in $\widehat{A}$.  

So suppose that for some $i,j,n, X$ we have a $\Sigma^0_n$ formula $\psi_j$ defining $X$ from $A$ and $X = \phi_i(0^{\omega})$.  Now since $\hat{T}_{\langle i,j \rangle}$ is arithmetic we can find an arithmetic path $Z$ through $\hat{T}_{\langle i,j \rangle}$ that's $n$-generic under local forcing.  Thus, the set defined by $\psi_j$ applied to $Z$ is also arithmetic and by construction must be equal to $\phi_i(0^{\omega}) = X$ (or we'd have found a disagreement).  Hence, $X$ is arithmetic (i.e. computable from $0^{m}$ for some $m$).