(a) If $f : D_2 \to D_2$ is a map such that $f(x) = x$, for $x \in S_1$, then there exists an interior point $z$ in the disk (a point $z ∈ D_2 − S_1$) such that $f(z) = z$.
(b) If $f : D_2 \to D_2$ is a map such that $f(x) = x$, for $x \in S_1$, then $f$ is surjective.
$(a)$ is false, $(b)$ is true.
There is a well-know theorem saying that there is no continuous map $f:D^2\to S^1$ so that $f(x)=x$ for any $x\in S^1$. (this is because $S^1$ has non-trivial fundamental group.)
This implies immediately $(b)$. Indeed, if $f$ is not surjective, there is $z\in D^2$ which is not in the image of $f$. Now, define $F(x)$ as follows. $F(x)$ is the intersection with $S^1$ of the line passing trhougt $z$ and $f(x)$, on the side of $f(x)$. $F$ would then be a continuous function from $D^2\to S^1$. Contradiction.
A counterexample of $(a)$.
Let $p\in S^1$ and consider the family of circles in $D^2$ which tanget to $S^1$ at $p$. By using a stereographic projection $\pi$ with pole $p$, the disk $D^2$ is transformed to the halfplane $\{(x,y)\in\mathbb R^2 :\ y\geq 0\}$, $S^1$ is mapped to the $x$-axis ($p$ is mapped to $\infty$) and the circles tangent at $p$ to horizontal lines $\{(x,y)\in\mathbb R^2 :\ y=c\}$
Define $f(x,y)=(x+y,y)$. Clearly, the fixed points of $f$ are just the points where $y=0$, that is to say, the points corresponding to $S^1$.
Thus the map $F=\pi^{-1}f\circ\pi:D^2\to D^2$ has the property that has no fixed point inside $D^2$ and that $F(x)=x$ for any $x\in S^1$.