In normal usage, is equality simply a context-dependent equivalence relation?

Short answer: Yes, you can. And this is what working mathematicians actually do, in the vast majority of cases, using a sensible abuse of notation to avoid pedantry.

Long answer: In the following, I'll be pedantic on purpose, to justify the fact that we can get rid of pedantry and, under appropriate conditions, lift the equivalence relation $\equiv$ to an equality $=$. To avoid too abstract generality, let's see a concrete example. Consider the construction of $\mathbb{Z}$ as the partition set generated by the following equivalence relation on $\mathbb{N} \times \mathbb{N}$: $(a,b) \equiv (c,d)$ if and only if $a+_\mathbb{N} d = c +_\mathbb{N} b$.

In such a context, we can define a function $+_\mathbb{Z}' \colon \mathbb{N} \times \mathbb{N} \to \mathbb{N} \times \mathbb{N}$ (acting on the elements of $\mathbb{N} \times \mathbb{N}$) such that \begin{align} (a,b) +_\mathbb{Z}' (c,d) = (a +_\mathbb{N} c, b +_\mathbb{N} d). \end{align}

It is immediate to prove that $+_\mathbb{Z}'$ is compatible with $\equiv$, i.e. it preserves equivalences classes: if $(a,b) \equiv (a',b')$ and $(c,d) \equiv (c',d')$ then $(a,b) +_\mathbb{Z}' (c,d) \equiv (a',b') +_\mathbb{Z}' (c',d')$ (indeed, $a + c + b' + d' = a' + c' + b + d$). Therefore, it is natural to lift $+_\mathbb{Z}'$ to a function $+_\mathbb{Z} \colon \mathbb{Z} \to \mathbb{Z}$ (acting on the equivalence classes of $\mathbb{N} \times \mathbb{N}$) defined by \begin{align} [(a,b)] +_\mathbb{Z} [(c,d)] = [(a+_\mathbb{N} c,b+_\mathbb{N} d)]. \end{align} Is this operation well-defined? Said differently, if we take different representatives of the same equivalence class, do we get the same result (as an equivalence class)? The compatibility condition above gives us a positive answer, i.e. the definition of $+_\mathbb{Z}$ does not depend on the choice of the representative of the equivalence classes: if $[(a,b)] = [(a',b')]$ and $[(c,d)] = [(c',d')]$ then $[(a,b)] +_\mathbb{Z} [(c,d)] = [(a',b')] +_\mathbb{Z} [(c',d')]$. Thus, it is perfectly legitimate to collapse $+_\mathbb{Z}'$ and $+_\mathbb{Z}$.

Suppose now that you only work on $\mathbb{N} \times \mathbb{N}$ with functions and relations defined on $\mathbb{N} \times \mathbb{N}$ that respect the compatibility condition. In such a context, it is impossible to distinguish $(1,0)$ and $(2,1)$, and more in general two distinct representatives of the same equivalence class. Therefore, in such a context, it is perfectly legitimate to collapse $\equiv$ (the equivalence relation on $\mathbb{N} \times \mathbb{N}$) and $=$ (seen as the equality on $\mathbb{Z}$), in accordance with Leibniz's logical principle of identity of indiscernibles: if two objects have the same properties, then they are identical. The fact that $(1,0)$ and $(2,1)$ are distinct ordered pairs does not matter as long as your are in a context where you cannot appreciate this difference, i.e. as long as you work only with functions and relations on $\mathbb{N} \times \mathbb{N}$ that are compatible with the equivalence relation $\equiv$.

Dully, given an equivalence relation $\sim$ on $X$, you can just define its unique "equality" relation $=_\sim$ on $X$ to be $x=_\sim y\iff x\sim y$. If you interpret every notion of equality as arising in this sense, then yes equality is simply a context-dependent equivalence relation.

For example, let $X$ be a set of words in English and define $\sim$ on $X$ by $x\sim y$ iff they have the same length. Then the words 'game' and 'dome' are $\sim$-equal in the above sense but are not literally syntactically equal.