The answer is no.
Here is a brief explanation: Consider a function $f$ of unbounded variation, where the increasing part comes from a differentiable function whereas the decreasing part comes from a "Cantor-staircase-type" function. Since $\int f'$ can only capture the increasing part, which itself is unbounded, the resulting integral diverges to $+\infty$ and hence $f'$ is not Henstock–Kurzweil integrable.
The following construction is a fleshed-out version of this idea.
Consider a sequence $(x_k)_{k\geq 0}$ such that $0 = x_0 < x_1 < x_2 < \cdots < 1$ and $\lim_k x_k = 1$. Also, for each $k\geq 1$, we pick a point $c_k \in (x_{k-1}, x_k)$. Then define the function $f : [0, 1] \to \mathbb{R}$ as follows:
On each $[x_{k-1}, c_k]$, the graph of $f$ is a line joining $(x_{k-1}, 0)$ to $(c_k, \frac{1}{k})$.
On each $[c_k, x_k]$, the graph of $f$ is a Cantor staircase joining $(c_k, \frac{1}{k})$ to $(x_k, 0)$.
$f(1) = 0$.
For instance, the figure below illustrates the graph of $f$ when $x_k = \frac{k}{k+1}$ and $c_k$'s are chosen as the middle points of the subintervals.
The first two condition guarantees that $f$ is continuous on $[0, 1)$. Also, since $0 \leq f(x) \leq \frac{1}{k}$ on each interval $[x_{k-1}, x_k]$, it follows that $f(x) \to 0$ as $x \to 1^-$ and hence $f$ is also continuous at $1$. It is also obvious that $f$ is differentiable almost everywhere with the a.e.-derivative
$$ f'(x) = \begin{cases}
\frac{1}{k(c_k - x_{k-1})}, & \text{if $x \in [x_{k-1}, c_k]$}, \\
0, & \text{elsewhere.}
\end{cases} $$
Since
$$ \lim_{k\to\infty} \int_{0}^{x_k} f'(x) \, \mathrm{d}x
= \lim_{k\to\infty} \sum_{j=1}^{k} \frac{1}{j}
= \infty, $$
it follows that $f'$ is not Henstock–Kurzweil integrable.
Not exactly sure of the formulation you mean, but the basic result here is that if $\alpha$ monotonic on $[a,b]$ then $\alpha$ is differentiable ae and $\int_a^b\alpha(x)dx \le \alpha(b)-\alpha(a)$ with the famous Cantor function a typical example when $\alpha'=0$ ae so the integral is zero but $\alpha(1)=1, \alpha(0)=0$ so RHS is $1$
(if $x$ is in the Cantor set so it has a decimal representation $x=0.x_1..x_n.., x_k =0,2$, then $\alpha(x)=0.y_1..y_n..$ where $y_k=x_k/2$, while if $x$ is in an excluded interval, first, it's easy to show that at the two ends which are in the Cantor set, $\alpha$ defined above takes the same value and then $\alpha(x)$ is defined as the common value at the two ends)
The differentiability ae of $\alpha$ monotonic is a difficult result but the integral inequality follows easily from Fatou's lemma.
We assume wlog that $\alpha$ is non decreasing and we extend it to the right of $[a,b]$ a little as the constant $\alpha(b)$ respectively; then since for each $h>0$ small $\frac{\alpha(x+h)-\alpha(x)}{h} \ge 0$ while the limit as $h \to 0$ is $\alpha'(x)$ almost everywhere (namely at the points of differentiability of $\alpha$), Fatou's lemma says that $$\liminf_{h \to 0^+}\int_a^b\frac{\alpha(x+h)-\alpha(x)}{h}dx \ge \int_a^b\alpha'(x)$$
But now for small enough $h>0$, $$\int_a^b\frac{\alpha(x+h)-\alpha(x)}{h}dx=\frac{\int_{a+h}^{b+h}\alpha(x)dx-\int_a^b \alpha(x)dx}{h}=$$ $$=\alpha(b)-\frac{1}{h}\int_a^{a+h}\alpha(x)dx$$
Since $\alpha$ is nondecreasing $\frac{1}{h}\int_a^{a+h}\alpha(x)dx \ge \frac{1}{h}\int_a^{a+h}\alpha(a)dx=\alpha(a)$,
so $\alpha(b)-\alpha(a) \ge \alpha(b)-\frac{1}{h}\int_a^{a+h}\alpha(x)dx$ and letting $h \to 0$ we are done!
Edit later: just to make the above clear, it may happen that $\alpha$ is monotone and continuous and the inequality is still strict as in the Cantor function case; we have equality (a non trivial result though) if $\alpha'$ exists everywhere (though it may not be continuous) and is integrable (which here holds automatically since by monotonicity $\alpha'$ has constant sign where it exists and by Fatou, its integral is finite on any finite interval) which then implies $\alpha$ absolutely continuous, and of course if $\alpha$ is absolutely continuous regardless of monotonicity (when $\alpha'$ may exist only ae)
Best Answer
The Cantor function allows you to go up or down without being accounted by the derivative.
Let $g$ be the Cantor function and define $$ f(x)=\begin{cases} x,&\ 0\leq x \leq \frac12\\[0.2cm] 1-g(x),&\ \frac12<x \leq1 \end{cases} $$ Then $f$ is continuous, it is differentiable almost everywhere, $f(1)=f(0)$, and $$ \int_0^1f'=\int_0^{1/2}1=\frac12. $$