Does Turing determinacy imply full determinacy?

This is open. In $L(\mathbb R)$ the answer is yes. Hugh has several proofs of this, and it remains one of the few unpublished results in the area. The latest version of the statement (that I know of) is the claim in your parenthetical remark at the end. This gives determinacy in $L(\mathbb R)$ using, for example, a reflection argument.

(I mentioned this a while ago somewhere on this site. Maybe that's where you heard of it? This can be used to prove that $\omega$-board determinacy is equiconsistent with determinacy. I seem to recall that's how the topic came up.)

To show this in L(R) (or any well understood determinacy model) one apparently has to run a core model induction as in section 6.2 of https://ivv5hpp.uni-muenster.de/u/rds/core_model_induction.pdf which explains why it is open in the abstract.