For typed calculi, if you consider the negative types ($1$, $\times$, $\to$), you can turn the eta rules on or off basically at will without affecting confluence.
For positive types (sums, and pairs with the pattern-matching elimination), the situation is much messier. Basically, the question is whether the term has a closed-scope elimination form, which let contexts interact in complicated ways with eta-expansions. For example, if $e$ has type $A \times B$, then its eta-expansion is $\mathsf{let}\;(a,b) = e \;\mathsf{in}\; (a,b)$. But to get the equational theory a category theorist would expect, you need to consider contexts $\mathbb{C}[-]$, and generalize the equation to be $\mathbb{C}[e] \equiv \mathsf{let}\;(a,b) = e\;\mathsf{in}\;\mathbb{C}[(a,b)]$ (with the expected scoping restrictions).
I think that you can still prove a confluence result if you don't allow the commuting conversions. But this is hearsay -- I've never tried it myself, nor looked at papers documenting it.
I don't really know anything about untyped lambda calculus, though.
EDIT: Charles asks about eta-reductions. This is promising for the kind of example he seeks, because I think in general they won't be strong enough to model the full equality theory, which I'll illustrate with a simple example involving booleans. The eta-expansion for booleans is $\mathbb{C}[e] \mapsto \mathsf{if}(e, \mathbb{C}[\mathsf{true}], \mathbb{C}[\mathsf{false}])$. (The eta-reduction is of course the other direction.)
Now, consider the term $\mathsf{if}(e, f, g)\;\mathsf{if}(e, x, y)$. Showing that this term is equivalent to $\mathsf{if}(e, f\;x, g\;y)$ needs to go through an eta-expansion, because we have to replace the $e$ in one of the if-then-elses with $\mathsf{true}$ and $\mathsf{false}$ in order to drive a $\beta$-reduction.
