In one variable, you can define the integral $\int_\gamma f(z)\,dz$ by $$\int_a^bf(\gamma(t))\cdot \gamma'(t)\,dt$$
You can approximate it by a Riemann sum
$$\sum_{k=0}^{n-1} f(z_k)(z_{k+1}-z_k)$$
If you have a primitive function $F(z)$, you can compute the integral as $F(b)-F(a)$, where $a$ and $b$ are the end points of the curve $\gamma$.
In several variables, all this fail because of dimensional considerations. Then $\gamma'(t)$ and $z_{k+1}-z_k$ are vectors instead of complex numbers, and there is no notion of a primitive function without specifying which variable you are differentiating with respect to. Hence the expression $\int_\gamma f$, integrating a function along a curve, is not even well defined. This is why we integrate 1-forms instead of functions. In contrast to the situation in one variable, there is no universal 1-form to use here.
Example
In ${\bf C}^2$, use the coordinates $(z,w)$ where $z=x+iy$ and $w=u+iv$. Consider the closed curve $\gamma$ given by $x^2+u^2=1$. Note that $\gamma$ is contained in ${\bf R}\times {\bf R}\subset {\bf C}\times {\bf C}$, and that we have no complex structure on ${\bf R}\times {\bf R}$. We can compute $\int_\gamma f(z,w)\,dz$ and $\int_\gamma f(z,w)\,dw$, but they reduce to ordinary real line integrals, and there is no reason why Cauchy's integral theorem should hold here. For instance, if $f(z,w)=w$ and we use the parametrization $z=\cos(t)$, $w=\sin(t)$, we get
$$\int_\gamma f(z,w)\,dz=\int_\gamma u\,dx = -\int_0^{2\pi}\sin^2(t)\,dt=-\pi$$
Best Answer
Yes, we can (assuming that $D$ is an open set). Let $z\in D$ and take $\varepsilon>0$. I will prove that there is a $\delta>0$ such that$$\lvert h\rvert<\delta\implies\left\lvert\frac{F(z+h)-F(z)}h-f(z)\right\rvert<\varepsilon.$$Let $r>0$ be such that $\overline{B(z,r)}\subset D$. Take $\delta>0$ such that $\delta<r$ and that $\lvert w-z\rvert<\delta\implies\bigl\lvert f(w)-f(z)\bigr\rvert<\varepsilon$. Then, for each $h$ such that $\lvert h\rvert<\delta$, consider the path $\gamma\colon[0,1]\longrightarrow\mathbb C$ defined $\gamma(t)=z+th$. Then\begin{align}\left\lvert\frac{F(z+h)-F(z)}h-f(z)\right\rvert&=\left\lvert\frac{\int_\gamma f(w)\,\mathrm dw-hf(z)}h\right\rvert\\&=\left\lvert\frac{\int_\gamma f(w)-f(z)\,\mathrm dw}h\right\rvert\\&<\frac{\varepsilon\lvert h\rvert}{\lvert h\rvert}\\&<\varepsilon.\end{align}