Note that the processes $$X(t,\omega)^+ := \max\{0,X(t,\omega)\} \qquad \qquad X(t,\omega)^- := \max\{0,-X(t,\omega)\}$$ are predictable and
$$\int_0^t X(r) \, dr = \int_0^t X_r^+ \, dr - \int_0^t X_r^- \, dr$$
Both integrals on the right-hand side are increasing in $t$, thus $t \mapsto \int_0^t X(r) \, dr$ is of bounded variation.
Suppose that $X$ has càdlàg sample paths. Let $\omega \in \Omega$, $T>0$ and $$c := \sup\left\{\int_0^s \left|X(r,\omega)\right| \, dr ; s \in [0,T]\right\}<\infty.$$
Then, by the mean value theorem,
$$\begin{align*} \left|\exp \left(\int_0^t X_r \, dr \right) - \exp \left( \int_0^s X_r \, dr \right) \right| = \exp(\xi) \cdot \left| \int_0^t X_r \, dr - \int_0^s X_r \, dr \right| \end{align*}$$
for any $s,t \in [0,T]$ and some $\xi=\xi(\omega) \in [-c,c]$. Thus
$$ \left|\exp \left(\int_0^t X_r \, dr \right) - \exp \left( \int_0^s X_r \, dr \right) \right| \leq e^c \cdot \left| \int_0^t X_r \, dr - \int_0^s X_r \, dr \right|$$
Consequently, the claim follows from the fact that $[0,T] \ni t \mapsto \int_0^t X(r) \, dr$ is of bounded variation.
I believe your argument provides a good interpretation for total (not quadratic) variation, because it is the one associated to the first derivative.
For quadratic variation, one should note first that squaring an infinitesimal element will cause it to shrink even more. So if you are adding up smaller things than before and it still results in a great number, then you can infer that the original increments were not that small. It might be an indication that a trajectory is very chaotic (a lot of randomness accumulated). In other words: the higher the quadratic variation, the higher is the amount of randomness.
Conversely, if a trajectory is "minimally smooth", although not differentiable, then the squaring procedure applied to the infinitesimal increments would be enough to keep the resulting summation controlled (sufficiently small). So you could also think about quadratic variation as a measured of "smoothness" of a trajectory.
Best Answer
In general if $X$ is a semimartingale and $H$ is a locally bounded predictable process, then $$ \Delta \left(\int_0^{\cdot} H_s\,\mathrm dX_s\right)_t=H_t \Delta X_t,\quad t\geq 0, $$ so if $X$ is continuous, then so is any integral with respect to $X$. Now, integration with respect to the Lebesgue measure is just integration with respect to the semimartingale $X_t=t$ (which actually is of finite variation). Since $\Delta X_t=t-t=0$ we have that this $X$ is continuous and hence the integral is as well.
The integral is also of finite variation because the following holds for the stochastic integral:
This can be seen in Jacod and Shiryaev's Limit Theorems for Stochastic Processes for example.