We consider sums of the form
$$\sum_{i=1}^m\gamma(t_i^*)\wedge\Big(\gamma(t_i)-\gamma(t_{i-1})\Big),$$
where $\gamma:[0,1]\to\mathbb R^n$ is a continuous function, and
$$0=t_0\leq t_1^*\leq t_1\leq t_2^*\leq t_2\leq\cdots\leq t_{m-1}\leq t_m^*\leq t_m=1$$
is a tagged partition of the unit interval, and $\wedge$ is the exterior product. As the norm of the partition $\max_i|t_i-t_{i-1}|\to0$, do these sums converge?
In particular (taking $n=2$), if $x,y:[0,1]\to\mathbb R$ are the coordinates of a plane curve, does
$$\int\big(x\,dy-y\,dx\big)$$
exist? This ought to be twice the area between the origin and the curve. Note, we can't break this into two integrals
$$\left(\int x\,dy\right)-\left(\int y\,dx\right)$$
because these may not exist. Continuous function need not be Riemann-Stieltjes integrable if $\alpha$ is not monotone.
Maybe a space-filling curve would be a counter-example.
As suggested by @user8268 (setting $a=2$ and replacing $t$ with $\sqrt t$), and as a modification of the linked answer, take the spiral curve
$$x=\sqrt t\sin1/t,\quad y=\sqrt t\cos1/t.$$
Let the tag points be $t_i^*=t_i$ to simplify:
$$x_i^*\big(y_i-y_{i-1}\big)-y_i^*\big(x_i-x_{i-1}\big)=x_{i-1}y_i-x_iy_{i-1}$$ $$=\sqrt{t_it_{i-1}}\sin\left(\frac1{t_{i-1}}-\frac1{t_i}\right).$$
(By continuity, when $t=0$ also $x=y=0$, so the first term in the Riemann sum is $x_0y_1-x_1y_0=0$.)
Suppose the norm of the partition is $\delta>0$, so $t_{i-1}\geq t_i-\delta$. To satisfy this restriction, fix the right half of the partition, where $t\geq\sqrt{2\delta/\pi}$, and for the left half of the partition take $t=2/(k\pi)$ for integers $k$ (in a finite but increasing range, including at least some $k\geq2/(\pi\delta)$ in consideration of the partition point $t=0$). So for $t\leq\sqrt{2\delta/\pi}$ we have
$$\frac1{t_{i-1}}-\frac1{t_i}=\frac\pi2$$ $$t_i-t_{i-1}=\frac\pi2t_it_{i-1}\leq\frac\pi2t_i^2\leq\delta$$
as required. The right half of the Riemann sum is fixed, and the left half is
$$\sum_k\sqrt{\frac2{k\pi}\frac2{(k+1)\pi}}\sin\left(\frac\pi2\right)$$ $$\geq\sum_k\frac2{(k+1)\pi}$$
which can be made arbitrarily large. Hence, the integral doesn't exist.
Surprisingly, even if the curve is on the unit circle, the integral (which ought to be proportional to the winding number) may not exist.
$$x=\cos\phi(t),\quad y=\sin\phi(t)$$ $$\sum_{i=1}^m(x_{i-1}y_i-y_{i-1}x_i)=\sum_{i=1}^m\sin\big(\phi(t_i)-\phi(t_{i-1})\big)$$
Such a sum is just what defines $\int_0^1\sin(d\phi(t))$. Since $d\phi$ is small (by uniform continuity of $\phi$, as the partition's norm is small), this integral ought to be $\int_0^1d\phi(t)=\phi(1)-\phi(0)$. In other words we should have $\int_0^1(\sin(d\phi(t))-d\phi(t))=0$. But it is not. Considering $\sin\theta-\theta\approx-\tfrac16\theta^3$, I was led to ask whether $\int_0^1(d\phi(t))^3=0$, and found a counter-example, which will also work here:
$$\phi(t)=\frac4{3\sqrt3}\sqrt[3]t\Big(\sin\tfrac13\tau/t-\tfrac12\sin\tfrac23\tau/t\Big).$$
Again using reciprocals of integers as partition points, three terms in the sum will be
$$\sin\left(\phi\left(\frac1{3k}\right)-\phi\left(\frac1{3k+1}\right)\right)+\sin\left(\phi\left(\frac1{3k+1}\right)-\phi\left(\frac1{3k+2}\right)\right)+\sin\left(\phi\left(\frac1{3k+2}\right)-\phi\left(\frac1{3k+3}\right)\right)$$ $$=\sin\left(0-\frac1{(3k+1)^{1/3}}\right)+\sin\left(\frac1{(3k+1)^{1/3}}-\frac{-1}{(3k+2)^{1/3}}\right)+\sin\left(\frac{-1}{(3k+2)^{1/3}}-0\right)$$ $$=\sin\left(\frac1{(3k+1)^{1/3}}\right)\left(-1+\cos\left(\frac1{(3k+2)^{1/3}}\right)\right)+\left(-1+\cos\left(\frac1{(3k+1)^{1/3}}\right)\right)\sin\left(\frac1{(3k+2)^{1/3}}\right)$$ $$=-\frac12\sin\left(\frac1{(3k+1)^{1/3}}\right)\sin^2\left(\frac12\frac1{(3k+2)^{1/3}}\right)-\frac12\sin^2\left(\frac12\frac1{(3k+1)^{1/3}}\right)\sin\left(\frac1{(3k+2)^{1/3}}\right),$$
and since $\sin\theta\geq\tfrac12\theta$ when $\theta>0$ is small enough, this is
$$\leq-\frac1{64}\frac1{(3k+1)^{1/3}}\frac1{(3k+2)^{2/3}}-\frac1{64}\frac1{(3k+1)^{2/3}}\frac1{(3k+2)^{1/3}}$$ $$\leq-\frac1{64}\frac1{(3k+2)^{1/3}}\frac1{(3k+2)^{2/3}}-\frac1{64}\frac1{(3k+2)^{2/3}}\frac1{(3k+2)^{1/3}}$$ $$=-\frac1{32}\frac1{(3k+2)}.$$
Thus the sum is unbounded below.
What's happening geometrically is that we're inscribing tiny obtuse triangles in the circle, and somehow making their areas add up.
The unit square is no better. Using the same partition points and function $\phi$ from the previous section, let
$$(x,y)=\begin{cases}(1,\;1+\phi(t)),\quad\phi(t)\leq0\\(1-\phi(t),\;1),\quad\phi(t)\geq0\end{cases}$$
so three terms $x_{i-1}y_i-y_{i-1}x_i$ in the sum will be
$$(1)\cdot\left(1+\frac{-1}{(3k+2)^{1/3}}\right)-(1)\cdot(1)\\+(1)\cdot(1)-\left(1+\frac{-1}{(3k+2)^{1/3}}\right)\cdot\left(1-\frac{1}{(3k+1)^{1/3}}\right)\\+\left(1-\frac{1}{(3k+1)^{1/3}}\right)\cdot(1)-(1)\cdot(1)$$ $$=-\frac1{(3k+1)^{1/3}(3k+2)^{1/3}}$$ $$\leq-\frac1{(3k+2)^{2/3}}$$
and again the sum is unbounded below, since $\sum_k1/k^{2/3}=\infty$.