Does $f'(x)>0$ a.e. imply that $f$ is strictly monotone? Let us assume that $f:\mathbb{R}\to \mathbb{R}$ is differentiable and $f'(x)>0$ almost everywhere. 
If $f'\in L^1_{loc}$, then FTC implies that for any $x,a\in \mathbb{R}$,
$$
f(x)-f(a)=\int_a^x f'(t)dt.
$$
Therefore, we have $f(x)\geq f(a)$, because $\int_a^x f'(t)dt\geq \frac{1}{n} m(\{x;f'(x)>\frac 1 n\})$ for any $n\in \mathbb{N}$. Then we know that $f$ has to be strictly monotone. Otherwise, there exits an interval $(c,d)\subset [a,b]$ such that $f(x)$ is constant over $(c,d)$, which is impossibe because $f'(x)=0$ is zero over this interval and hence $f'(x)>0$ a.e. fails. 
How about removing the assumption that $f'\in L^1_{loc}$? Does $f'(x)>0$ almost everywhere imply that $f$ is strictly monotone
 A: This is a special case of the Goldowsky-Tonelli theorem. See Saks, Theory of the integral, Chapter 6, page 206. (Thanks to https://mathoverflow.net/a/266399/112284 for the reference)

Your argument applies verbatim without the $L^1_{loc}$ restriction using the Henstock-Kurzweil integral. For this integral, the FTC holds for any differentiable function, it isn't affected by sets of zero measure, and the integral of an a.e. positive function is strictly increasing.

Here is another proof that avoids the Henstock-Kurzweil integral. First, note it suffices to show weak monotonicity, $f(b)\geq f(a)$ for all $b>a$ - we would then only have equality if the function was constant in $[a,b],$ which would give a derivative of zero.
Apply the Vitali-Carathéodory theorem to get an upper semicontinuous function $g:[a,b]\to\mathbb R$ satisfying $g(x)\leq \min(0,f'(x))$ for all $x\in[a,b]$ and $\int_a^b g(t)dt\geq -\epsilon.$ Define $h(x)=f(x)-\int_{a}^x g(t)dt+(x-a)\epsilon.$ Using upper semicontinuity, $\limsup_{y\to x}\frac{1}{y-x}\int_x^y g(t)dt\leq g(x).$ We get $\liminf_{\substack{y\to x\\ y\in[a,b]}}\frac{h(y)-h(x)}{y-x}\geq f'(x)-g(x)+\epsilon\geq \epsilon$ for all $x\in [a,b].$ It follows that $h(b)\geq h(a)$: consider the infimal $x$ such that $h(x)<h(a)$; it must satisfy $h(x)=h(a),$ but since the lower (lim inf) derivatives are positive, we must have $h(y)>h(a)$ for sufficiently small $y-x>0.$ This gives $f(b)\geq f(a)-\epsilon(1+b-a),$ and since $\epsilon$ was arbitrary, $f(b)\geq f(a).$
