The Cauchy–Riemann equations and analyticity

No.

Let $c$ be the Cantor function on $[0,1]$, so that $c$ is continuous, $c' = 0$ a.e., but $c$ is not constant. Then take $u(x+iy) = v(x+iy)=c(x)c(y)$. We have $u_x=u_y=v_x=v_y=0$ a.e. so the Cauchy–Riemann equations are trivially satisfied, and $f(z)=u(z)+iv(z)$ is bounded and continuous on the unit square, but certainly not analytic.

Almost everywhere differentiability is almost never the right condition for solutions to a PDE. A better condition would be to have $u,v$ in some Sobolev space.

See this related question. Denjoy proved that there exist a continuous function $f$ on the unit square and a continuous curve $\gamma$, which is the graph of a continuous function, such that $f$ is holomorphic on the square minus $\gamma$ but not on the whole square. Thus $f$ satisfies the CR equations almost everywhere, and actually on the whole square minus the support of $\gamma$, but not on the whole square.

Thus the general answer to your question seems to be a solid no. Using the postive parts of Denjoy's result, one can imagine to answer in the affirmative if the set where CR fails is a countable union of curves with sufficiently nice behaviour, but it seems difficult to do better than Sindalovskii. See also here for a different proof of his result.