This isn't quite an answer but it's almost a reduction to a famous unsolved problem. Consider the question of whether there is an $\omega$-REA set of minimal arithmetic degree. If this question has a relativizing solution (e.g. either there is an $\omega$-REA operator A(X) which produces a minimal cover for X in the arithmetic degrees or for all such A, X A(X) is not a minimal arithmetic cover) then it's answer decides this question.

If there is such a minimal cover operator then, by the inversion theorem for $\omega$-REA operators there is an operator J s.t. A(J(0)) is Turing equivalent to $0^{\omega}$ and J(0) non-arithmetic so it is minimal cover. OTOH the operator A(X) that always returns $X \oplus 0^{\omega}$ would necessarily yield a minimal cover for some $X$ if $0^{\omega}$ was a minimal cover.

This isn't quite an answer but it's almost a reduction to a famous unsolved problem. Consider the question of whether there is an $\omega$-REA set of minimal arithmetic degree. If this question has a relativizing solution (e.g. either there is an $\omega$-REA operator A(X) which produces a minimal cover for X in the arithmetic degrees or for all such A, X A(X) is not a minimal arithmetic cover) then it's answer decides this question.

If there is such a minimal cover operator then, by the inversion theorem for $\omega$-REA operators there is an operator J s.t. A(J(0)) is Turing equivalent to $0^{\omega}$ and J(0) non-arithmetic so $0^{\omega}$ is a minimal cover. OTOH the operator A(X) that always returns $X \oplus 0^{\omega}$ would necessarily yield a minimal cover for some $X$ if $0^{\omega}$ was a minimal cover.