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 $n \to \infty$.
The embedding above of $V$ into $H$ is not compact, but this can be fixed, as follows. Take $D$ to be the infinite diagonal matrix with $n$th diagonal entry equal to $1/n$. Set $A = D^{1/2} H D^{1/2}$. $A$ is compact as an operator on $V = \ell^2$, and can be taken to define the inner product for a Hilbert space in which $V$ is compactly embedded. The analysis for the corresponding projection $Q_n = V_n (V_n^* A V_n)^{-1} V_n^* A$ proceeds as before, $u_i$ now given by $-(\tfrac{i}{n+1})^{3/2}$. The squared norm of the projection in this case is then $$ \Vert Q_n \Vert^2 = \tfrac{n}{4}+\tfrac{3}{4}+\tfrac{1}{4}\tfrac{1}{n+1}, $$ which still diverges to infinity.