Ohh, I think I'm being dumb. The answer is no.

Given $X \not\leq_T 0'$ we build $Y$ using the finite extension method and modify the usual minimal pair construction by coding in the bits of $X$ into $Y$ between the minimal pair requirements.

Now, since 0' can figure out how the minimal pair requirements are met it can decode the bits of $X$ in $Y$. Thus $0' +Y$ computes $X$ but any $Z$ must be computable and as $X \not\leq_T 0'$ the claim fails.

Ohh, I think I'm being dumb. The answer is no.

Given $X \not\leq_T 0'$ we build $Y$ using the finite extension method and modify the usual minimal pair construction by coding in the bits of $X$ into $Y$ between the minimal pair requirements.

Now, since $0'$ can figure out how the minimal pair requirements are met it can decode the bits of $X$ in $Y$. Thus $0' +Y$ computes $X$ but any $Z$ must be computable and as $X \not\leq_T 0'$ the claim fails.

Ohh, I think I'm being dumb. The answer is no.

We build $X$ and $Y$ using the finite extension method and modify the usual minimal pair construction by also insisting that $X$ isn't $0'$ computable (we just diagnolize against the computations from 0' in between meeting the usual minimal pair requirements) and we code in the bits of $X$ into $Y$ between the minimal pair requirements.

Now, since 0' can figure out how the minimal pair requirements are met it can decode the bits of $X$ in $Y$. Thus $0' +Y$ computes $X$ but any $Z$ must be computable and as $X \not\leq_T 0'$ the claim fails.

Ohh, I think I'm being dumb. The answer is no.

Given $X \not\leq_T 0'$ we build $Y$ using the finite extension method and modify the usual minimal pair construction by coding in the bits of $X$ into $Y$ between the minimal pair requirements.

Now, since 0' can figure out how the minimal pair requirements are met it can decode the bits of $X$ in $Y$. Thus $0' +Y$ computes $X$ but any $Z$ must be computable and as $X \not\leq_T 0'$ the claim fails.