Let $z = r(\cos \alpha + i\sin \alpha), \qquad w = r(\cos \beta + i\sin \beta)$ since $|z|=|w|=r$ and both are nonzero, so $r>0$.

We are given $\operatorname{Arg} z + \operatorname{Arg} w = \pi$ which means $\alpha + \beta = \pi.$ Hence, $\alpha = \pi - \beta.$

Substitute into $z$: $z = r\big(\cos(\pi-\beta)+i\sin(\pi-\beta)\big).$ Using identities, $\cos(\pi-\beta)=-\cos\beta, \qquad \sin(\pi-\beta)=\sin\beta.$ So, $z = r(-\cos\beta+i\sin\beta).$

Now find $\overline w$: $w = r(\cos\beta+i\sin\beta)$ so $\overline w = r(\cos\beta-i\sin\beta).$ Therefore, $-\overline w = r(-\cos\beta+i\sin\beta).$

Comparing with the expression for $z$, $z=-\overline w.$

Hence the correct option is: $\boxed{\text{B}}$