This type of intermediate connectivity is often called semi-connectivity.
If you used this notion you would find out that your question already has a very comprehensive answer on stackoverflow: https://stackoverflow.com/questions/30642383/determine-if-a-graph-is-semi-connected-or-not
I will sketch it here simplified though.
Idea is to reduce problem by reducing digraph G to it's graph of SCC's, say G', that is always acyclic and then try to find a path that covers all nodes in G'. If there is no such path then it's not possible to get from some SCC to some other (G' is acyclic) and that means that vertices in them have no connections. Otherwise G' is semi-connected and so is G.
Input: digraph G = (V,E)
Output: true or false if it's semi-connected
init graph G'=(V',E') from SCC's of G where V' is set of SCC of G and E' are edges between them or $E'=\{(v_1',v_2') | v_1',v_2'\in V'$ and $\exists (v_1,v_2)\in E$, $v_1 \in v_1',v_2 \in v_2'$}
// G' guaranteed to be acyclic and this can be done in O(|V| + |E|), for example, using Kosaraju's algorithm
find topological ordering $T=(v_0,...,v_{|V'|})$ of G' // correct as G' is acyclic and also can be done in O(|V'| + |E'|) using modification of DFS.
if some possible pair $(v_i,v_{i+1})$ in T is not in E' ouput false, otherwise true // if not then $(v_{i+1},v_{i})$ also is not in E' and there is no connection between vertices in $v_i$ and $v_{i+1}$, takes O(|V'|)
Any $n$-vertex graph with average degree at most $2$, no matter what the maximum degree is, has a vertex cover of size at most $\frac23n$. Also, if the average degree is exactly $2$, then the number of edges is also $n$, and this gives the bound you wanted.
To see this, start from the Caro-Wei theorem, which guarantees that in any graph $G$, there is an independent set of size at least
$$
\sum_{v \in V(G)} \frac1{\deg(v) + 1}.
$$
By convexity of $x \mapsto \frac1{x+1}$ (for nonnegative $x$) this is at least $\frac{n}{d+1}$, where $d$ is the average degree. (This statement is also a variant of Turán's theorem.)
If the average degree is at most $2$, then there is an independent set of size at least $\frac13n$, and its complement is a vertex cover of size at most $\frac23 n$.
Best Answer
Yes, this is true for all connected graphs on more than $1$ vertex. (Or, for an even weaker assumption, all graphs with no isolated vertices.)
Let $S$ be a maximal independent set. Each vertex of $S$ must be adjacent to some vertex of $V \setminus S$, since otherwise it wouldn't be connected to anything.
This gives us at least $|S|$ edges incident to $|V\setminus S|$ vertices, so one of the vertices must be incident to at least $\frac{|S|}{|V\setminus S|}$ of them.