HI/P reduction mechanism
According to the source mentioned in the comments to your question, the first step is indeed nucleophilic substitution of the OH group by $\ce{I-}$, faciliated by protonation of the alcohol. For the second step ($\ce{HI}$ reduction), a radical species was found as an intermediate, and therefore a reduction by single electron transfer (SET) with oxidation of $\ce{I-}$ to $\ce{I2}$ is proposed. Red phosphorus is required for the regeneration of $\ce{HI}$ from $\ce{I2}$. $\ce{P_{red}}$ reacts with $\ce{I2}$ to the phosphorus iodides $\ce{PI3}$ and $\ce{PI5}$, which are subsequently hydrolyzed to $\ce{H3PO3}$/$\ce{H3PO4}$ and $\ce{HI}$, which is reused in further reduction steps. Catalytic amounts of $\ce{HI}$ are therefore sufficient to run the reaction in the presence of phosphorus. However, a too low concentration of $\ce{I-}$ will lead to elimination instead of substitution in the first reaction step.



(Image source)

The SET mechanism of the reduction is not described in detail in the article, so the following is a bit speculative. As a resonance-stabilized radical is involved, homolytic cleavage of the $\ce{C-I}$ bond is one of the first steps, which gives the benzylic radical $\ce{R.}$ and an iodine radical $\ce{I.}$. A single electron is transferred from $\ce{I-}$ to $\ce{R.}$, forming the carbanion $\ce{R-}$ and $\ce{I.}$, which recombines with another iodine radical to $\ce{I2}$. Protonation of $\ce{R-}$ yields the final product.