This isn't true. E.g. when $A$ is the negative of the identity matrix, its characteristic polynomial is $(x+1)^3=x^3+3x^2+3x+1$.
In general, the characteristic polynomial of a $3\times3$ nonsingular matrix $A$ is in the form of
\begin{aligned}
&x^3-\sum_i\lambda_i(A)x^2+\sum_{i\ne j}\lambda_i(A)\lambda_j(A)x-\prod_i\lambda_i(A)\\
=\,&x^3-\operatorname{tr}(A)x^2+\det(A)\operatorname{tr}(A^{-1})x-\det(A).
\end{aligned}
You are essentially asking whether
$$
\operatorname{tr}(A)=\det(A)\operatorname{tr}(A^{-1})
$$
when $A$ is orthogonal. Since $A^{-1}=A^T$ in this case, the above equality is true if and only if $\operatorname{tr}(A)=0$ or $\det(A)=1$.