If you assume that $b(t;\mathbf{v},\mathbf{v}) \approx (\mathbf{v},\mathbf{v})_H$ for every $\mathbf{v}\in H$, and if you assume that the vector $\mathbf{u} = \sum u_i(t) w_i \in H$, then Igor's answer goes through essentially unchanged in the infinite dimensional case.

But let me give you a counterexample if the assumptions are not verified.

Let the index $i$ run from $0, 1, \ldots$.

Let $b(t; w_i, w_j)$ be given by the matrix
$$ \begin{pmatrix}
1 & -1 \\
-1 & 2 & -1 \\
& -1 & 2 & -1 \\
&& -1 & 2 & -1 \\
&&& \ddots & \ddots & \ddots\end{pmatrix} $$

Assume that $w_i$ forms an orthonormal basis of $H$. We have that
$$ b(\mathbf{v}, \mathbf{v}) = v_0^2 + 2 \sum_{i = 1}^{\infty} v_i^2 - 2 \sum_{i = 0}^{\infty} v_{i} v_{i+1} = \sum_{i = 0}^\infty (v_i - v_{i+1})^2 $$
Since the right hand side is non-negative, we see that the bilinear form is positive definite, since
$$ b(\mathbf{v},\mathbf{v}) = 0 \implies v_i = v_{i+1} $$
which for $\mathbf{v} \in H$ requires $v_i \equiv 0$.
So $b(t) = b$ give inner products.

But the same argument above also shows that for any constant $c$,
$$ u_i(t) \equiv c \quad \forall i $$
is a solution to the infinite system of ODEs
$$ \partial_t u_i = \sum b(t;w_i,w_j) u_j $$
and so you don't have uniqueness of your solution. Note, of course, the vector $\mathbf{u} = \sum u_i w_i \not\in H$ whenever $c \neq 0$.