I was thinking about this question while walking home today and can't seem to prove or come up with a counterexample myself.
Let $f:[a,b]\rightarrow\mathbb{R}$ be a continuous function, $f(x),$ differentiable everywhere on $(a,b)$, except for one point $\xi$ in $(a,b)$ where $f'(\xi)$ does not exist. Since $f$ is continuous everywhere and differentiable everywhere on $(a,b)$ except at $\xi$, $f'$ is gauge integrable on $[a,b]$. Am I correct that $\int_a^bf'(x)\;\mathrm{d}x=f(b)-f(a)$ despite that nasty point $\xi$ in $(a,b)$? And could this value be interpreted as the "signed" area under the curve from high school calculus?
If $f(x)=x\sin (\frac 1 x)$ ($f(0)=0$) then $f$ satisfies your hypothesis but $f$ does not satisfy the Fundamental Theorem of Calculus . If it did satisfy this on each interval contained in $[-1,1]$ then it would be absolutely continuous and hence of bounded variation on $[-1,1]$ but it is not of bounded variation.