Let $J$ be an $*$-automorphism on a II$_1$-factor with a faithful normal tracial state $\tau$. Since $J$ preserves the identity, it follows that $\tau(J\cdot)$ is also a faithful normal tracial state. By the uniqueness of this tracial state, we obtain that $\tau(J\cdot )=\tau(\cdot)$. Hence, $J$ preserves the traces.
For $B(h)$, all automorphisms are inner, and therefore, they preserve traces.
I would like to ask whether the statement is also true for the II$_\infty$ factor? That is, can I say that all $*$-automorphisms on a II$_\infty$ factor preserves traces.
No. Take for instance $M=R\otimes B(H)$, with $R$ the hyperfinite II$_1$-factor. Normalize the trace $\tau$ so that $\tau(1\otimes E_{11})=1$. We can write $M$ as $M=R\otimes M_2(\mathbb C)\otimes B(H)$. Let $\theta$ be an isomorphism between $R\otimes M_2(\mathbb C)$ and $R$, and define $J=\theta\otimes\operatorname{id}$. Consider $p=1\otimes (E_{11}+E_{22})\otimes E_{11}$. Then $Jp$ is the identity in $R$, and $$ \tau(p)=2,\qquad \tau(Jp)=1. $$