First of all, $H_*(T^2,D^2)$ is not isomorphic to $H^*(T^2\setminus D^2)$. By the long exact sequence of the pair $H_*(T^2,D^2)$ it is possible to show that $H_*(T^2,D^2)\cong \tilde{H}_*(T^2)$, where this last notation denotes reduced homology. On the other hand $T^2\setminus D^2$ is homotopy equivalent to the one-point union of two circles. In particular $H_2(T^2\setminus D^2)=0$, but $H_2(T^2,D^2)=\mathbb Z$.
Excision can tell you that $H_*(T^2,D^2)\cong H_*(T^2\setminus int(D^2),\partial D^2)$. This is because we are excising $int(D^2)$. (Recall excision give an isomorphism $H_*(X,Y)\cong H_*(X\setminus U,Y\setminus U)$. Actually, to apply excision, the closure of $U$ needs to be contained in the interior of $Y$, so you need to fuss a little with deformation retracts first.)
(Also, perhaps you meant $H_*(T^2,D^2)\cong \tilde{H}_*(T^2/D^2)$, where $T^2/D^2$ is the quotient space? This is true.)