Theorem. The map $V(A,H)\to\operatorname{Hom}(A,\mathbb{B}(H))$ is open.
We write $\omega_{\xi,\eta}$ for the linear functional $x\mapsto \langle x\xi,\eta\rangle$ and $\omega_\xi$ for $\omega_{\xi,\xi}$. It is an elementary fact that if $\omega_\xi=\omega_{\eta}$ on a von Neumann algebra $A$, then $a\xi\mapsto a\eta$ extends to a partial isometry $u\in A'$ such that $u\xi=\eta$. We can trim $u^*u$ a little, if necessary, and make it satisfy $1-u^*u \sim 1-uu^*$ (Murray--von Neumann equivalence), at the cost of $\|u\xi-\eta\|<\epsilon/2$, where $\epsilon>0$ is arbitrary small. Then $u$ extends to a unitary element in $A'$, still denoted by $u$, which satisfies $\|u\xi-\eta\|<\epsilon$. The following perturbation lemma is well-known and follows from the theory of standard form (see [Takesaki, Section IX]).
Lemma.
For any $\epsilon>0$, there is $\delta>0$ which satisfies the following.
For any von Neumann algebra $A\subset B(K)$ and any unit vectors
$\xi,\eta\in K$, if $\|(\omega_\xi-\omega_\eta)|_A\|<\delta$,
then there is a unitary element $u\in A'$ such that
$\|u\xi - \eta\|<\epsilon$.
We postpone the proof of this lemma and prove the theorem. Let $v_0\in V(A,H)$ and an SOT neighborhood $$G=\{ v\in V(A,H) : \forall i\ \|(v-v_0)\xi_i\|<\epsilon\}$$ be given. Here $\xi_1,\ldots,\xi_n\in H$ are unit vectors and $\epsilon>0$. Take $\delta>0$ from the Lemma for $n^{-1/2}\epsilon$. Now suppose $v\in V(A,H)$ is such that $$\| (\omega_{\xi_i,\xi_j}\circ\operatorname{Ad}_v - \omega_{\xi_i,\xi_j}\circ\operatorname{Ad}_{v_0})|_M \|<\delta/n$$ for all $i,j$. We consider the unit vector $\xi=n^{-1/2}\left[\begin{smallmatrix} \xi_1 & \cdots & \xi_n\end{smallmatrix}\right]^T\in H^n$ and view $\mathbb{B}(H^n)=\mathbb{M}_n\otimes\mathbb{B}(H)$. Then $$\|(\omega_{(1\otimes v)\xi} - \omega_{(1\otimes v_0)\xi})|_{\mathbb{M}_n\otimes A}\|<\delta.$$ Thus by Lemma, one finds a unitary element $u\in A'\cong (\mathbb{M}_n\otimes A)'\cap\mathbb{B}(H^n)$ such that $\|(1\otimes u)(1\otimes v)\xi - (1\otimes v_0)\xi\|<n^{-1/2}\epsilon$. This implies that $uv\in G$, which finishes the proof.
Proof of Lemma. We may assume $K = p(L^2A \otimes \ell_2)$, where $L^2A$ is a standard representation of $A$ and $p\in (A\otimes \mathbb{C}1)'\cap\mathbb{B}(L^2A\otimes\ell_2)$. Fix a unit vector $\delta_0\in\ell_2$. There are unique vectors $|\xi|$ and $|\eta|$ in the positive cone $(L^2A)_+$ such that $\omega_\xi=\omega_{|\xi| \otimes\delta_0}$ and $\omega_\eta=\omega_{|\eta| \otimes\delta_0}$ on $A \otimes \mathbb{C}1$ (see [Takesaki, Theorem IX.1.2.(iv)]). There are partial isometries $v$ and $w$ in $(A\otimes \mathbb{C}1)'\cap\mathbb{B}(L^2A\otimes\ell_2)$ such that $\xi=v|\xi|$ and $\eta=w|\eta|$. By the generalized Powers--Stormer inequality ([Takesaki, Theorem IX.1.2]), one has $\| |\xi| - |\eta| \|^2 \le \|(\omega_\xi-\omega_\eta)|_A\|$. Hence $t:=wv^*\in p(A\otimes \mathbb{C}1)'p = A'\cap \mathbb{B}(K)$ satisfies $$\| t \xi - \eta \| = \| w(v^*\xi-w^*\eta)\|\approx 0.$$ Let $t=u|t|$ be the polar decomposition. Since $\|t\|\le1$ and $\|t\xi\|\approx\|\eta\|=1$, one has $|t|\xi \approx \xi$ and $u\xi\approx\eta$. We can further replace the partial isometry $u\in A'$ with a unitary element without affecting $u\xi$ much.