Ok, I'm pretty sure the answer is no. 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 $0^{\omega +1}$
Now the resulting set $A$ is clearly non-arithmetic and arithmetic in $0^{\omega}$ and if $\widehat{A} \leq_a A$ with $\widehat{A} \leq_T 0^{\omega}$ then for some $i,j,n$ we have a $\Sigma^0_n$ formula $\psi_j$ defining $\widehat{A}$ from $A$ and $\widehat{A} = \phi_i(0^{\omega})$. 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 generic and by construction must be equal to $\phi_i(0^{\omega}) = \widehat{A}$. Hence, $\widehat{A}$ is arithmetic and hence $A$ can't be arithmetic in $\widehat{A}$.