I think the answer, in general, is no. Here's a counter-example; perhaps someone else can find a simpler one. Take $V = \ell^2$, and let the inner product for $H$ restricted to $V$ correspond to the infinite tridiagonal matrix
$$
\begin{bmatrix} 2 & -1 & & \\ -1 & 2 & -1 & \\ & \ddots & \ddots & \ddots \end{bmatrix}.
$$
I'll call this matrix $H$ so that $(u,v)_H = (Hu,v)$ for $u,v \in \ell^2$. Now let $V_n$ just be the span of the first $n$ standard basis vectors. Let me abuse notation and also use $V_n$ for the matrix with these vectors as columns. Then
$$
  Q_n = V_n (V_n^* H V_n)^{-1} V_n^* H
$$
This infinite matrix is zero outside of the $n \times n+1$ upper-left block, which is given by $\begin{bmatrix} I_n & u \end{bmatrix}$ with $u_i = -\tfrac{i}{n+1}$. The square of the $\ell^2$ norm of $Q_n$ works out to be $1+\Vert u \Vert^2$, which is
$$
  \Vert Q_n \Vert^2 = \tfrac{n}{3} + \tfrac{5}{6} + \tfrac{1}{6} \tfrac{1}{n+1},
$$
and so $\Vert Q_n \Vert \to \infty$ as $n \to \infty$.