Why is it meaningless to say the runtime of an algorithm is at least in the order of n squared?
First of all, both sentences are a slight abuse of notation. $\mathcal{O}(n^2)$ and $\Theta(n^2)$ are sets; strictly speaking you can't compare a running time with a set. Saying "$f$ is $\mathcal{O}(g(n))$" is actually a shorthand for "$f$ belongs to the set $\mathcal{O}(g(n))$"; with that in mind, we can rephrase the statements you quote as follows:


  The running time of $A$ is at least $f(n)$ for some $f \in \mathcal{O}(n^2)$.


and:


  The worst-case running time of $A$ is at least $f(n)$ for some $f \in \Omega(n^2)$.


Put in this form, the first statement is well-defined, but it's not really insightful. In fact, it is always true. Let $g$ be the constant function $0$. Certainly $g$ is a member of $\mathcal{O}(n^2)$, and whatever the running time of $A$ is on input of size $n$, it is at least $g(n)$.

Intuitively speaking, big-oh is an upper bound: if I tell you: "I have at least $100 or less", I didn't really give you any information at all.

On the other hand, the second statement guarantees that the worst-case running time of $A$ cannot be too low: it must grow at least quadratically.