Let $\mathbb D^n$ be the closed unit disk in $\mathbb R^n$. Let $f:\mathbb D^n \to \mathbb{R}^n$ be a smooth immersion . ($df$ is everywhere invertible). Let $\omega:\mathbb D^n \to \mathbb{R}^n$ be the harmonic map corresponding to the Dirichlet problem imposed by $f$, i.e. $\omega|_{\partial \mathbb D^n}=f|_{\partial \mathbb D^n}$, and each component of $\omega$ is a harmonic function.
Does $\text{rank}(d\omega)\ge n-1 $ everywhere on $\mathbb D^n$?
For every $p \in \partial \mathbb D^n :\, \, \text{rank}(d\omega_p)\ge \text{rank}\big(d(\omega|_{\partial \mathbb D^n})_p\big)= \text{rank}\big(d(f|_{\partial M})_p\big)=n-1$.
Since the rank does not decrease locally, there is an open neighbourhood of $\partial \mathbb D^n$ where $\text{rank}(d\omega)\ge n-1 $.
Edit: In the particular case of dimension $2$, the question is whether $\omega$ has a critical point ($d\omega=0$).
Let us consider the 2$D$ case more closely: Write $f=(f_1,f_2)$ for $f_i:\mathbb D^2 \to \mathbb R$, and similarly $\omega=(\omega_1,\omega_2)$. Then $df_i \neq 0$ everywhere on $\mathbb D^2$. However, this fact alone does not imply that $d\omega_i \neq 0$ on $\mathbb D^2$. A counter example is given here:
Set $f_i(x,y)=x-2y^2$. Then $\omega_i(x,y)=x^2-y^2+x-1$, and $d\omega_i=(2x+1,-2y)$ is zero at the point $(-\frac{1}{2},0)$.
Of course, to get a counter-example to the current question, we need $d\omega_1,d\omega_2$ to vanish at the same point simultaneously; However we cannot take $f_1=f_2$, since then $f=(f_1,f_2)$ won't be an immersion. I tried tweaking this construction, e.g. by composing one copy of the $f_i$ with some rotation around $(-\frac{1}{2},0)$, but this got me nowhere.