As Hsieh notes, your definition of conductance is off from the one I know by a factor of $d$, where $d$ is the degree of the regular graph. This is also known as edge expansion for regular graphs.
A relationship between edge expansion and diameter is quite easy to show. Intuitively, an expander is "like" a complete graph, so all vertices are "close" to each other. More formally, let
$$
\min_{S \subseteq V}\ { \frac{ e(S, S^c) }{ d \cdot \min\{|S|, |S^c|\} }} \geq \alpha
$$
Take any set of vertices $S$ with $|S| \leq |V|/2$. There are at least $\alpha d |S|$ edges coming out of $S$ and since $G$ is $d$-regular, the neighborhood of $S$ (including $S$ itself) is of size at least $(1+\alpha)|S|$. Applying this claim inductively, starting from $S = \{u\}$ for any vertex $u$, we see that for some $t = O(\log_{1 + \alpha } |V|)$, $u$'s $t$-hop neighborhood has size at least $|V|/2$. Therefore, the $t+1$-hop neighborhood of any vertex $v$ has to intersect the $t$-hop neighborhood of $u$, or the graph would have more than $|V|$ vertices, a contradiction. So you have
$$
D = O\left(\frac{\log |V|}{\log (1 + \alpha)}\right)
$$
Of course, it also follows that having a lower bound on the diameter implies an upper bound on edge expansion.
I don't think small diameter implies conductance. If you don't insist on regular graphs (and use Hsieh's definition), then two complete graphs connected by a single edge provides a counterexample.
