Assume $f(x)$ is an $n$ bits to $m$ bits function and we want to use Grover's search algorithm to find $x$ such that $f(x) = a$, where $a$ is some $m$-bit predetermined value.
When using the algorithm, can we separate the calls to $f(x)$ and $f(x)=a$? That is, instead of querying $F(x) \equiv (f(x) = a)$ around $2^{n/2}$ times (interleaved with amplification), is it possible to query $f(x)$ once at the begining and then repeat $2^{n/2}$ times $f(x)=a$ (interleaved with amplification) ?
It sounds like you want a state given to you as $\vert x\rangle\vert f(x)\rangle$, and you wish to only conditionally rotate when $f(x)=a$, diffuse over $x$, conditionally rotate when $f(x)=a$, diffuse over $x$, ... conditionally rotate when $f(x)=a$, diffuse over $x$, and measure.
If I understand your question, then I think the issue is in the "diffuse over $x$."
Because you can't simply diffuse over the mixed state $\vert x\rangle$, but you must diffuse over the pure state $\vert x\rangle\vert f(x)\rangle$.
If $\vert f(x)\rangle$ has not been uncomputed back to $\vert 0\rangle$, then the diffusion will not lead to constructive/destructive interference in the first register. Even if you are diffusing only over the first register $\vert x\rangle$, the second register $\vert f(x)\rangle$ gets in the way.
