Maximal class for which function equivalence is decidable
There is no such maximal class.

Proof: suppose that there is a class $C$ of representations of total functions for which function equivalence is decidable. If $C$ is finite, then $C$ augmented with the constant function that returns $\max\{\overline{f_i}(0)\} + 1$ (which by construction is not in $C$) has decidable function equivalence (test the function at $0$ then apply the postulated decision algorithm). If $C$ is infinite, it is denumerable (because representations of functions are denumerable¹), so let $(f_i)_{i\in\mathbb{N}}$ be an enumeration of all function representations in $C$. Let $G(i) = \overline{f_i}(i) + 1$ (a classic diagonal construction). By construction, $G$ does not have a representation in $C$ (because $\forall i, g_i(i) \ne h(i)$). Let $g$ be a representation of a function that calculates $G$ (this is doable in any Turing-equivalent framework). The algorithm that reports that $g$ is distinct from any member of $C$, and otherwise applies the existing decision algorithm for $C$, is an equivalence decision procedure for $C \cup \{g\}$.

Any class of function representations with decidable equivalence can be augmented to a larger such class. Therefore there is no maximal class.

Intuitively speaking, if you have two classes of functions with decidable equivalence, you can apply the two decision algorithms inside each class, but it's not necessarily possible to compare two functions from different classes. And if you have an infinite set of classes, with a decision procedure for each, then even calculating which decision procedure to use is not necessarily doable.

¹  The set of all functions is not denumerable, but whatever representation of functions you choose (e.g. source code in some language), the set of inputs for your decision algorithm is denumerable.