Well, the title almost says it all. I would like to list as many examples as possible of moduli functors, for which a coarse moduli space does not exist (and maybe explain why). So, examples such as $[\mathbb{A}^1/G_m]$ are not what I mean. I would really like to see moduli functors that come from some classifying problem.

$\overline{\mathfrak{M}}_{0,0}$, the stack of prestable genus zero curves. This parameterizes genus zero curves having at worst nodal singularities. There are an infinite number of isomorphism classes of such curves, but they all occur as a specialization of a trivial family (just do repeated blowups of the central fiber of the family $\mathbb{P}^1\times \mathbb{A}^1 \to \mathbb{A}^1$ ). 


There is an example in HarrisMorrison Moduli Of Curves (Exercise 1.7): the moduli functor $F:\mathfrak{Sch}/\mathbb C\to\textrm{Sets}$ that sends a scheme $B$ to the set of isomorphism classes of (flat $B$families of) reduced plane curves of degree $2$. The point is to show that there exists a "universal" natural tranformation $\eta:F\to \hom_{\mathbb C}(,\textrm{Spec }\mathbb C)$. So by the uniqueness of the moduli solution, $\textrm{Spec }\mathbb C$ is the unique reasonable candidate as a moduli space for this problem. But unfortunately $F(\textrm{Spec }\mathbb C)$ has two $\mathbb C$points (the class of $\mathbb P^1$ and the class of a double line), while $\textrm{Spec }\mathbb C$ has only one. It's likely that one can also find an example where a classifying space exists, but is not universal. 


Here is a classifying problem without a coarse moduli space: Let $F$ be the stack of line bundles with section, say over an algebraically closed field k. That is $F(X)$ is the category of pairs $(L,s)$ where $L$ is a line bundle on $X$ and $s \in \Gamma(X,L)$ is a section of $L$. Then $F$ has no coarse moduli space: if it did, there would be two points, corresponding to $(k,1)$ and $(k,0)$ with the former specializing to the latter. This would be an affine scheme with a closed point and a generic point, both with residue field $k$. That's impossible. Of course, $F$ is just a geometric description of $[\mathbf{A}^1 / \mathbf{G}_m]$, so we already knew it didn't have a coarse moduli space. 

