I know I am about a decade late to the party, but I am about to start teaching an undergraduate complex analysis course in a couple of days and have been trying to think about a geometric interpretation of the contour integral to help explain to people whose mathematical background is at the basic vector calculus level. Here is one thing I have come up with, that I think is more or less "dual" to the answer by @Dirk. I'll be curious to see if anyone finds it helpful.
If we write $z=x+iy$, then we can write a complex valued function as $f(z)=u(x,y)+iv(x,y)$. A slightly different point of view is that the function $f(z)$ defines a vector field on $\mathbb{R}^2$ given by $f(x,y)=\langle u(x,y),v(x,y)\rangle$. If $\gamma(t)=x(t)+iy(t)$ is a contour then by unpacking the definition of $\int_\gamma f(z)dz$ and breaking into real an imaginary parts we get a pair of line integrals:
$$\int_\gamma f(z)dz=\int_\gamma udx-vdy+i\left(\int_\gamma vdx+udy\right)=\int_\gamma \langle u,-v\rangle\cdot ds+i\int_\gamma \langle v,u\rangle \cdot ds,$$
where $\tilde ds=\langle dx,dy\rangle$ and $ds$ is the corresponding unit tangent vector to $\gamma$. In other words, $ds$ is the line element tangent to the curve $\gamma$. By a standard trick from vector calculus, we can rewrite a 2-dimensional line integral as a flux integral of a different vector field. More precisely, let $\tilde dn=\langle -dy,dx\rangle$ and $dn$ the corresponding unit vector. We can think of $dn$ as an outward unit normal element to the curve $\gamma$. A simple computation give $\langle u,-v\rangle\cdot ds=\langle v,u\rangle \cdot dn$, and so
$$\int_\gamma \langle u,-v\rangle\cdot ds=\int_\gamma \langle v,u\rangle\cdot dn.$$
Thus the real part of $\int_\gamma f(z)dz$ is the flux of the vector field $\langle v,u\rangle$ through the curve $\gamma$. Returning to the complex setting we can interpret the vector field $\langle v,u\rangle$ as the complex function $g(z)=i\overline{f(z)}$.
A similar computation for the imaginary part of $\int_\gamma f(z)dz$ shows that it is equal to the flux of the vector field $\langle -u,v\rangle$ through the curve $\gamma$. On the complex side we see that this vector field corresponds to the function $h(z)=-\overline{f(z)}$.
Note that $h(z)=-ig(z)$ and so these two function/vector fields just differ by rotation by $\pi/2$ With this perspective the contour integral $\int_\gamma f(z)dz$ comes from combining the fluxes of $g(z)$ and its rotation $h(z)$ through the curve $\gamma$.
If $f(z)$ is analytic this says that $\overline{f(z)}$ and the "rotated" function $i\overline{f(z)}$ both have zero flux through any closed curve.
If $f(z)=1/z$ then this says that $-\frac{1}{\bar{z}}$ thought of as a vector field on $\mathbb{R}^2$ has flux $2\pi$ through any curve that winds once around 0 and that the rotated vector field $\frac{i}{\bar{z}}$ has zero flux through any curve that winds once around 0.