Universal Quantifiers in QBFs
Your intuition was right.  It doesn't work.  Here is a counterexample.
Consider $\forall a \exists b  \varphi(a,b)$ where $\varphi(a,b) = (a \lor \neg b) \land (\neg a \lor b)$.  This statement evaluates to true.
However, if we remove $a$ following your procedure, we get $\exists b \psi(b)$ where $\psi(b) = ((\neg b) \land (b))$; and that statement evaluates to false.
One way to understand why your method doesn't work is that $\forall a \exists b  \varphi(a,b)$ isn't equivalent to $\exists b \forall a  \varphi(a,b)$.  At most your method might work if all of the $\forall$'s are on the inside, i.e., for a sentence of the form $\exists \cdots \exists \forall \cdots \forall$, but not for any other pattern of $\exists$ and $\forall$.