Yes. Let $g$ be an antiderivative of $f'_{app}$, then $(f-g)'_{app}=0$ , this implies that $h:=f-g=\text{const}$.

Sketch of the proof of this fact. It suffices to prove that $h(x)=h(0)$ a.e. Indeed, if $h(a)\ne 0$, than this would imply $h'_{app}(a)=\infty$. We may fix $c>0$ and $\rho\in (0,1)$ and prove that the measure of $x\in [0,1]: h(x)-h(0)<cx$ is at least $\rho$. This, again, suffices. We say that a point $t\in [0,1]$ is nice if $\mu\{x\in [0,t]:h(x)-h(0)\geqslant cx\}\leqslant (1-\rho)t$, and $t$ is very nice if $t$ is a density point for the set of nice points. Let $a$ be a supremum of very nice points in $[0,1]$. If $a<1$, we get $h(a)-h(0)\leqslant ca$, else $h'_{app}$ would be infinite. Next, there exist a segment $[a,a+\varepsilon]$ such that $h(x)-h(a)<c(x-a)$ for more than $\rho\cdot \varepsilon$ points in this segment. This allows to find a very nice point greater than $a$. A contradiction.