If you take a map $f$ of the form that you propose, with
$$\psi(r) = \frac{6}{5}r\left(1-\frac{1}{6}r^4\right),$$
and $c = 12/5$, then the ODE $(\psi' + r^{-1}\psi)^2 + (\phi'\psi)^2 = c^2$ for $\phi$ becomes
$$\phi'(r) = 2r(1-r^4/4)^{1/2}(1-r^4/6)^{-1}.$$
This has an analytic solution for $r < \sqrt{2}$ of the form
$$\phi(r) = r^2\left(1 + \sum_{k \geq 1} a_kr^{4k}\right).$$
The corresponding map
$$f(z) = \left(r^{-1}\psi e^{i \phi}\right)z = \frac{6}{5}z(1-|z|^4/6)e^{i |z|^2\left(1 + \sum_{k \geq 1} a_k|z|^{4k}\right)}$$
is analytic in $D$ and satisfies the desired conditions with $c = 12/5 > 2$. 

(Alternatively, choose $\psi$ to be any concave smooth increasing function with $\psi(0) = 0$ and $\psi(1) = 1$ such that $\psi$ is linear near the origin. Then $f$ is a just a dilation near the origin and we don't need to worry about singularities there).