Let $D^n$ be the closed unit ball in $\mathbb{R}^n$ and assume that for all smooth $\phi:D^n\rightarrow D^n$ and all $x\in D^n$ we have $\phi(x)\neq x$.
Then a function $F:D^n\rightarrow \partial D^n$, where $\partial D^n=\mathbb{S}^{n-1}$ is the boundary of $D^n$, with $$F(x)=\frac{x-\phi(x)}{|x-\phi(x)|}$$is smooth and well-defined.
Let $G:\mathbb{S}^{n-1}\rightarrow\mathbb{S}^{n-1}$ be the restriction of $F$ to $\mathbb{S}^{n-1}$.
Why is the degree of $G$ equal to deg$G=0?$
Since $G$ is homotopic to the identity on $\mathbb{S}^{n-1}$ and degrees are homotopy-invariant, we find a contradiction. And thus $f$ must have a fixed-point.
What I thought:
Let $\omega$ be a $(n-1)$-form on $\mathbb{S}^{n-1}$ s.t. $\int_{\mathbb{S}^{n-1}}\omega=1$. Then deg$G=\int_{\mathbb{S}^{n-1}}G^*\omega$.
How do I use Stokes to find that the degree is $0$?
If $\omega$ is a volume from on $S^{n-1}$ like you said, then degree of $G$ is $\displaystyle \int_{S^{n-1}} G^* \omega$. This $S^{n-1}$ is the boundary $\partial D^n \subset D^n$. By Stokes' theorem (as $G^*\omega = F^*\omega|_{\partial D^n}$):
$$\int_{\partial D^n} G^* \omega = \int_{D^n} d(F^*\omega) = \int_{D^n} F^*(d\omega) = 0$$
As $d\omega$, an $n$-form on $S^{n-1}$, is zero. That's the desired contradiction.