Trade off between time and query complexity for total functions
Here is a sketch of an attempt at partial separation. I was trying to improve this over the past two weeks into a full answer, but haven't made much progress. Hopefully someone else knows how to extend it.

Let 

\[f_{n,m}: \{0,1\}^n \times \{0,1\}^{{m \choose 2}} \rightarrow \{0,1\}\]

So we have a total function on tuples. 

$f_{n,m}(x,y)$ returns $1$ if $x = 1^n$ and if $y$ interpreted as the edges of a graph has a clique or independent graph of size $n$. Otherwise, it returns $0$. 

Thus if $R_n$ is the $n$-th diagonal Ramsey number then for $m \geq R_n$ and all $y \in \{0,1\}^{{m \choose 2}}$ we have $f(1^n,y) = 1$. Thus, if we can compute $R_n$ on an input of $1^n$ then we don't have to read the bits of $y$ if $m$ is large enough, and the query complexity of the query-optimal algorithm drops to $n$.

However, $R_n$ is harder to compute on an input of $1^n$ than it is to check if a graph given by edges $y$ has a clique of independent set of size $n$ on input $(1^n,y)$. Thus, a time-optimal algorithm can't afford to compute $R_n$. It could, instead use some approximation $\tilde{R}_n &gt; R_n$. For $m &lt; \tilde{R}_n$ a time-optimal algorithm will read at least some bits of $y$.

Thus, for $R_n \leq m &lt; \tilde{R}_n$ the query-optimal algorithm will use more time to make fewer queries than the time-optimal algorithm.

This approach can be made slightly less hand-wavy (or generalized to other Ramsey-like problems) but I think even the more rigorous attempt doesn't fully answer the question.