The answer is no.
Let be construct such a $(V,\sigma)$. Let $X$ be the set of functions $\mathbf{N}\to\mathbf{R}$. Let $Y\subset X$ be a subset such that
- $Y$ is linearly independent
- $Y$ contains all Dirac functions $n\mapsto\delta_{m,n}$
- for every $g\in X$, there exists $f\in Y$ such that $\limsup(f/(|g|+1))=\infty$.
Note that the last condition forces $Y$ to be uncountable. The existence of $Y$ is checked by considering a maximal linearly independent subset containing all Dirac functions.
Now consider the space $V$ with basis indexed by $\mathbf{N}\sqcup Y$, namely $(e_n)_{n\in\mathbf{N}}$ and $(z_f)_{f\in Y}$, and symmetric bilinear form $$\sigma(e_n,e_m)=\delta_{m,n},\quad \sigma(e_n,z_f)=f(n),\quad\sigma(z_f,z_g)=0.$$
I claim that $\sigma$ is non-degenerate; I'll use the first two axioms. Indeed, let $v=\sum_na_ne_n+\sum_fb_fz_f$ belong to the kernel of $\sigma$. Then $0=\sigma(v,z_f)=\sum_na_nf(e_n)=0$ for all $f\in Y$. Choosing $f=\delta_n$, we deduce $a_n=0$, for all $n$. So $v=\sum_fb_fz_f$. Then $0=\sigma(v,z_n)=\sum_{f\in Y}b_ff(n)$ for all $n$, which means that $\sum_{f\in Y}b_ff=0$. Since $Y$ is linearly independent, we deduce $b_f=0$ for all $f$, and hence $v=0$.
Now suppose the existence of $H$, $S:V\to H$ and $T$ as required. Write $E_n=S(e_n)$ and $Z_f=S(z_f)$, $C_f=\|TZ_f\|$, $g(n)=\|E_n\|$. Then for all $f\in Y$ and $n$, we have $f(n)=\sigma(z_f,e_n)=\langle TZ_f,E_n\rangle$. So $|f(n)|\le C_fg(n)$ for all $n$. Hence $\limsup (f/(g+1))<\infty$. Since this holds for each $f\in Y$, we contradict the third axiom, which ends the proof.
However, I don't have at the moment a counterexample for which $V$ has countable dimension.