Does the Brouwer fixed point theorem hold for the torus minus an open disk?

Let's express $T^2$ by letting $H \subset \mathbb{R}^2$ be a regular hexagon centered at the origin, and then gluing opposite sides of $H$ by translations.

Let $q : H \to T^2$ be the quotient map of this gluing.

Let $D^2 \subset \mathbb{R}^2$ be a small round open disc centered at the origin and contained in the interior of $H$, and so we may regard $D^2$ as embedded in $T^2$ by the map $q$. Then, as you say, we remove the interior of $D^2$ to obtain the surface $S$.

Now let $R : H \to H$ be a rotation of angle $2 \pi / 6$. The map $R$ induces a homeomorphism $S \mapsto S$ having no fixed points.

Let's sit your torus $T^2$ inside $\Bbb{C} \times \Bbb{C}$ as the subset $S^1 \times S^1$, where $S^1 = \{ z \mid |z|= 1\}$ and let's sit your disk $D^2$ so that it doesn't meet the circle $\{(z, 1) \mid z \in S^1\}$. Then the function $(z, w) \mapsto (iz,1)$ has no fixed points on any subset of $T^2$.