The dual cone would include non-symmetric matrices that give non-negative quadratic forms.
Follow the same proof. Assume that $q^TYq\geq0$ for all vectors $q$, but $Y$ not necessarily symmetric. Let $X$ a (symmetric) positive semi-definite matrix, i.e. any element of the set that we are computing the cone of. Write it as $X=\sum_i \lambda_i q_iq_i^T$, with $\lambda_i\geq0$.
Then $$tr(YX)=\sum_i \lambda_i tr(Yq_iq_i^T)=\sum_i \lambda_i tr(q_i^TYq_i)\geq0$$
Therefore, $Y$ belongs to the dual cone of the positive semi-definite matrices.
This shows that the dual cone of the (symmetric) positive semi-define matrices, when considered in the space of all matrices, will be strictly larger.