In a von Neumann algebra, if $A_{\alpha}$ converges to $0$ in the $\sigma$-weak topology, do the positive parts $(A_{\alpha})_{+}$ necessarily converge to $0$ in the $\sigma$-weak topology?

Nope, not even in the abelian case. Work in $L^\infty[0,1]$. Let $f_n$ be the function which is alternately plus and minus $1$ on the subintervals $[\frac{i}{2^n}, \frac{i+1}{2^n}]$. Then $f_n \to 0$ weak* but the positive parts converge to $1/2$.