Yes, the projection is $1$-Lipschitz. In the half-space model, choose $\alpha$ to be the vertical half-line $\{(0,0,\dots,t):t>0\}$. For each $t$,
the set of points that project to the point $P_t=(0,\dots,t)$ is the union of all geodesics leaving $P_t$
orthogonally to $\alpha$. These geodesics lie of the Euclidean sphere $S(t)=\left\{x:\sum_{k=1}^n x_k^2 = t^2\right\}$.
Now the problem boils down to showing that for any $0<t<s$ the distance between $S(t)$ and $S(s)$ is attained by the pair of points $P_t $ and $P_s $. Consider any path $\gamma:[0,1]\to \mathbb H^n$ that begins on $S(t)$ and
ends on $S(s)$. Write it in coordinates as $\gamma=(\gamma_1,\dots,\gamma_n)$, so that the length of $\gamma$ is $\int \frac{d\gamma}{\gamma_n}$. For comparison, consider the path
$$\tilde \gamma = \left(0,0,\dots,\sqrt{\sum_{k=1}^n \gamma_k^2}\right)$$
which connects $P_t$ to $P_s$. Clearly, $|\tilde \gamma'|\le |\gamma'|$ (Euclidean norms), and $\tilde \gamma_n\ge \gamma_n$.
Hence,
$$d(P_t,P_s)\le \int \frac{d\tilde \gamma}{\tilde \gamma_n}\le \int \frac{d\gamma}{\gamma_n}$$
and therefore $d(P_t,P_s)\le d(a,b)$ whenever $a\in S(t)$ and $b\in S(s)$.
