I hope it's okay to post an answer to my own question. I am essentially repeating Woodin's proof that he just showed me. Any errors were probably introduced by me.
Recall that under AD all measures on ordinals are countably complete and ordinaldefinable.
By the coding of measures theorem of Kechris assuming AD and that there is a Suslin cardinal above $\kappa$, there are fewer than $\Theta$ many measures on $\kappa$ (or $\kappa^{<\omega}$ for that matter.)
So we may push forward Martin's measure on $\mathcal{P}_{\omega_1}(\mathbb{R})$ to get a fine, countably complete measure $U$ on $\mathcal{P}_{\omega_1}(\operatorname{meas}(\kappa^{<\omega}))$, where $\operatorname{meas}(\kappa^{<\omega})$ is the set of all measures on $\kappa^{<\omega}$.
Fix a tree $T$ on $\omega\times \kappa$. As usual, given a real $x \in \omega^\omega$ we let $T_x = \lbrace s \in \kappa^{<\omega} : (x \restriction s, s) \in T\rbrace$.
Claim: $U$almost all $\sigma$ witness the weak homogeneity of $T$.
Proof: For each $\sigma$ we define a game $G_{\sigma}$, closed for Player I, for which Player I has a winning strategy iff $\sigma$ does not witness the weak homogeneity of $T$.

I plays: $(x_0, \alpha_0, \beta_0)$, $(x_1, \alpha_1,\beta_1),\ldots$

II plays: $\mu_0$, $\mu_1,\ldots$
Let $x$, $\vec{\alpha}$, $\vec{\beta}$, and $\vec{\mu}$ denote the resulting sequence of moves.

Rules for I: $\vec{\alpha} \in [T_x]$ and the sequence $\vec{\beta} \in \mathrm{Ord}^\omega$ continuously witnesses that the tower $\vec{\mu}$ is illfounded.

Rules for II: $\vec{\mu}$ is a tower of measures in $\sigma$ concentrating on $T_x$.
If both players follow the rules until the end, we say that player I wins.
If $\sigma$ does not witness that $T$ is weakly homogeneous, say $x \in p[T]$ but there is no wellfounded tower of measures in $\sigma$ concentrating on $T_x$, then there is a continuous witness to the illfoundedness of towers of measures in $\sigma$ concentrating on $T_x$. (This is proved by an argument similar to what follows but using a fine, countably complete measure on the set of subsets of $\kappa^{<\omega}$.)
So if $\sigma$ does not witness that $T$ is weakly homogeneous, then player I has a winning strategy in $G_\sigma$. (The $x$ and $\vec{\alpha}$ are fixed in advance and the $\vec{\beta}$ comes from $\vec{\mu}$ via the continuous witness mentioned above.) Assume toward a contradiction that for $U$almost every $\sigma$, player I has a winning strategy in $G_\sigma$. The game is closed, the moves are ordinals and measures, and all measures are ordinaldefinable, so for such $\sigma$ player I has a winning strategy $F(\sigma)$ of playing the least move leading to a subgame where he or she still has a winning strategy.

Define the integer $x_0$ to be the one played by $F(\sigma)$ on the first turn for $U$almost every $\sigma$.

Define a measure $\mu_0$ on $\kappa$ by $A \in \mu_0 \iff \forall^*_U \sigma\; (\alpha^{\sigma}_0 \in A)$ where $\alpha^{\sigma}_0$ is the ordinal $\alpha_0$ played by $F(\sigma)$ on the first turn.

Define the integer $x_1$ to be the one played by $F(\sigma)$ on the second turn against $\mu_0$ for $U$almost every $\sigma$.

Define a measure $\mu_1$ on $\kappa^2$ by $A \in \mu_1 \iff \forall^*_U \sigma\; ((\alpha^{\sigma}_0,\alpha^{\sigma}_1) \in A)$ where $\alpha^{\sigma}_1$ is the ordinal $\alpha_1$ played by $F(\sigma)$ against $\mu_0$ on the second turn.
Continuing in this way, we get a real $x \in \omega^\omega$ and a sequence of measures $\vec{\mu}$. One can easily check that $\vec{\mu}$ is a tower of measures.
Each $\mu_i$ concentrates on $T_x$ because $(\alpha_0^\sigma,\ldots,\alpha_i^\sigma) \in T_x$.
It is a wellfounded tower because if $A_i \in \mu_i$ for all $i<\omega$ then by countable completeness of $U$ there is a $\sigma$ such that $(\alpha_0^\sigma,\ldots,\alpha_i^\sigma) \in A_i$ for all $i<\omega$.
However, by countable completeness of $U$ there is a $\sigma$ such that $\vec{\mu}$ is a legal play by player II against player I's winning strategy $F(\sigma)$, so player I's moves $\beta^\sigma_i$ continuously witness the illfoundedness of $\vec{\mu}$. Contradiction.