I am working with real symmetric non-negative matrices with integer elements and zero diagonal. They are particularly nice, and I am sure that they are conditionally negative definite. That is, if $A$ is an $n\times n$ matrix just described, and $x$ is any $n$-vector such that $\sum_i x_i = 0$, then $$x^TAx \leq 0.$$ I know of Shoenberg's result stating that this is true if and only if there are vectors $y^1, \ldots, y^n \in \mathbb{R}^s$ for some $s$, such that $a_{ij} = ||y^i - y^j||^2$. In other words that square roots of the elements of $A$ express the distances between $n$ points in $s$ dimensions.
I have not yet been able to prove this generally for all $A$ however.
No. Take, eg, $A=\begin{pmatrix}0&1&16\\1&0&4\\16&4&0\end{pmatrix}$ and $x=(1,-2,1)^T$. Then $x^TA x = 12>0$. So $A$ is not conditionally negative definite.
That such examples exist follows from Shoenberg's theorem. Start with a function that is not a metric, such as the "side lengths" of a triangle with sides 1,2,4. Fill the matrix $A$ with the squares of these lengths. If it's conditionally neg. definite, there would be a triangle with sides of length 1, 2, and 4.