This is a bit of a brute force approach, but it's effective. Note that the sum of the reverse diagonal elements of a $2\times 2$ matrix $M$ equals ${\rm tr}\,\sigma M$ with
$$\sigma=\begin{pmatrix}0&1\\1&0\end{pmatrix}.$$
 For the most general form of the matrix
$$A=\begin{pmatrix}a&b\\ -b&c\end{pmatrix},\;\;\text{and for}\;\;D=\mathbf c\mathbf b^{\rm T}=\begin{pmatrix}1&-1\\1&-1\end{pmatrix},$$
I calculate
$$J(r,n)={\rm tr}\,\sigma A^r DA^{n-1-r}=$$
$$=\frac{2^{-n-1} (a+c-z)^{-r} (a+c+z)^{-r}}{(a-2 b-c) \left(a c+b^2\right)} \left[\left(z (a+c)-(a-c)^2+4 b^2\right) (a+c+z)^n (a+c-z)^{2 r}-\left(z (a+c)+(a-c)^2-4 b^2\right) (a+c-z)^n (a+c+z)^{2 r}\right],$$
with the definition $z=\sqrt{(a-c)^2-4 b^2}$.
Then I evaluate for $n\geq 1$ the sum
$$\sum_{r=0}^{n-1}J(r,n)=\frac{2^{-n-1} (a+c) \left((a-c)^2-4 b^2-z^2\right) \left((a+c-z)^n-(a+c+z)^n\right)}{z (a-2 b-c) \left(a c+b^2\right)}.$$
Substitution of the definition of $z$ finally gives the desired result
$$\sum_{r=0}^{n-1}J(r,n)=0.$$

--- 
<sub>
Details of the calculation: I may assume $b\neq 0$ (otherwise $A$ is diagonal and the identity follows trivially). Then the matrix $A$ is diagonalizable when $b\neq \tfrac{1}{2}|a-c|$, in the form
$A=U\Lambda U^{-1}$ with $$U=\left(
\begin{array}{cc}
 z-a+c & -z-a+c \\
 2 b & 2 b \\
\end{array}
\right)$$ $$\Lambda={\rm diag}\,\left(\tfrac{1}{2} \left(-z+a+c\right),\tfrac{1}{2} \left(z+a+c\right)\right)$$
With this decomposition we can readily evaluate $A^r=U\Lambda^r U^{-1}.

If $b=\tfrac{1}{2}(a-c)\neq 0$ we instead use the Jordan decomposition $A=VJV^{-1}$ with
$$V=\left(
\begin{array}{cc}
 -1 & -\frac{2}{a-c} \\
 1 & 0 \\
\end{array}
\right),\;\;J=\left(
\begin{array}{cc}
 \frac{a+c}{2} & 1 \\
 0 & \frac{a+c}{2} \\
\end{array}
\right).$$
Then $A^r=VJ^r V^{-1}$, with $J^r=2^{-r} (a+c)^r\left(
\begin{array}{cc}
 1 & 2 r (a+c)^{-1} \\
 0 & 1 \\
\end{array}
\right)$.
</sub>