I wrote a short paper to answer this question. Here's a summary. Let's shift $U$ so it contains $0$ and normalize the biholomorphic map $f : \mathbb D\to U$ by $f(0)=0$.

By the maximum principle, it suffices to have $|f'|\le 1$ near the boundary of $\mathbb D$, which can be expressed in terms of the harmonic measure of $U$ about $0$: its density must be at least $\frac{1}{2\pi}$ everywhere. Thinking of harmonic measure as the exit distribution of Brownian motion starting at $0$, we can see three ways in which it can have low density:

- Some part of the boundary is far away, hard to reach
- Some part of the boundary is too close to $0$, so it intercepts too many particles.
- Some part of the boundary is too curved, hard to squeeze in (Brownian motion has a hard time traveling through narrow corridors).

Let's assume $U$ is convex. With such a domain we associate three radii:

*Outer radius* $R_O$ is the smallest radius of a disk centered at $0$ and containing $U$
*Inner radius* $R_I$ is the largest radius of a disk centered at $0$ and contained in $U$
*Curvature radius* $R_C$ is the minimal radius of curvature of $\partial U$. (Equivalently, $R_C$ is the largest radius $R$ such that $U$ can be written as a union of open disks of radius $R$.)

Note that $R_O\ge R_I$ and $R_O\ge R_C$, while there is no general relation between $R_I$
and $R_C$.

For $a,b>0$ define
$$
\phi(a,b) = \begin{cases} \frac{\log a-\log b}{a-b} ,\quad & a\ne b \\ \frac{1}{a} ,\quad &a=b \end{cases}
$$

**Theorem**. If
$$
(R_O-R_C) \phi( R_I, R_C) + \frac12 \log R_C\le 0
\tag1$$
then a conformal map $f:\mathbb D\to U$ with $f(0)=0$ satisfies $|f'|\le 1$ in $\mathbb D$.

Remark: when $U$ is the disk of radius $R$, the left side of (1) is equal to $\frac12 \log R$, i.e., the requirement is $R\le 1$. So the condition (1) is sharp in some sense.

Idea of proof: for $z\in \mathbb D$, let $d=\operatorname{dist}(f(z), \partial U)$. It suffices to show $d(z)\le 1-|z|$ when $z$ is close to $\partial\mathbb D$. To do this, estimate the hyperbolic distance from $0$ to $f(z)$ from above, using the convexity of $U$ together with the existence of large disks in $U$. By the conformal invariance of the hyperbolic metric, a lower bound on it gives an upper bound on $|z|$, which leads to $|z|\le 1-d(z)$.