If $D$ is simply connected this follows from the Riemann mapping theorem.
If not, let's see. There exists $r>0$ such that $B(0,r)\subset D$. And there exists $M$ so $|z|\le M$ for all $z\in D$, hence $|f(z)|\le M$. So we have $|f(z)|\le M$ in $B(0,r)$, hence Cauchy's estimates show there exists $c$ such that $$|f'(0)|\le c$$for every holomorphic $f:D\to D$.
Now let $g=f\circ f$. Then $g'(0)=f'(f(0))f'(0)=f'(0)^2$. So in fact we have $$|f'(0)|\le c^{1/2},$$since $|g'(0)|\le c$. The same trick shows that $$|f'(0)|\le c^{1/n}$$for every $n$, hence $|f'(0)|\le 1$.