So the oxidation steps of the typical anitmarkovnikov Brown hydroboration/oxidation goes as shown in attached image A. Peroxide anion (not HO radical) attacks boron. Alkyl migration (with loss of HO-), followed by basic hydrolysis gives the antimarkovnikov alcohol.
So by analogy, here's what I think might (might) be happening in vivo (attached image B). Same attack of HOO-, alkyl migration, and hydrolysis gives a deprotonated phenol derivative. The alkoxide anions collapse to form the C=O double bond, and kick out the O- on the opposite side of the ring.
Can't promise it's correct, but I can push the arrows to make it work.