It is relatively easy to give an explicit sequence of inner automorphisms that converges to the flip automorphism $\sigma$. First of all, realize $R$ as an infinite tensor product of matrix algebras $(R,\varphi)=\bigotimes_n (M_{k_n}(\mathbb{C}),\varphi_n)$. For every $n\in \mathbb{N}$, we find a unitary $u_n\in M_{k_n}(\mathbb{C})\otimes M_{k_n}(\mathbb{C})$ that implements the flip automorphism on $M_{k_n}(\mathbb{C})\otimes M_{k_n}(\mathbb{C})$, i.e. $u_n(x\otimes y)u_n^\ast=y\otimes x$. (This follows from the general observation that every automorphism of a type I factor is inner, but it is a good exercise to find the $u_n$ explicitly).
Now it follows that the sequence of inner automorphisms $\sigma_n=Ad_{u_1\otimes\ldots\otimes u_n\otimes 1\ldots}$ converges to the flip automorphism $\sigma$: Let $\psi$ be any ultraweakly continuous functional on $R\otimes R$. Consider $R\otimes R$ to be represented on the infinite tensor product space $H=\bigotimes_n L^2(M_{k_n}(\mathbb{C})\otimes M_{k_n}(\mathbb{C}), \varphi_n\otimes\varphi_n)$. We know that $\psi$ is of the form $\psi(x)=\sum_k\langle\xi_k,x\eta_k\rangle$ for some $\ell^2$-summable sequences $\xi_k,\eta_k$ in $H$. Since the finite sequences are dense in the $ell^2$-summable ones, we can assume that $\psi(x)=\langle\xi,x\eta\rangle$. Because the finite tensor products are dense in the infinite ones, we can assume that $\xi,\eta\in H_N=\bigotimes_{n=1}^N L^2(M_{k_n}(\mathbb{C})\otimes M_{k_n}(\mathbb{C}), \varphi_n\otimes\varphi_n)$ for some $N\in\mathbb{N}$. Now it follows that $\psi(\sigma_n(x))=\langle \xi,\sigma_n(x)\eta\rangle=\langle\sigma_n^{-1}(\xi),x\sigma_n^{-1}(\eta)\rangle=\langle\sigma^{-1}(\xi),x\sigma^{-1}(\eta)\rangle=\psi(\sigma(x))$.
