No.  $C_{2k+1}$ is self-centered with $r(G)=d(G)=k$, but obviously does not contain $C_{2k}$ as a subgraph.  

With regards to the revised question, here is a proof that $G$ contains a cycle of length at least $2m$.  

*Proof.* Let $G$ be a connected graph with $r(G)=d(G)=m>1$.  I first claim that $G$ is 2-connected.  To see this, suppose that $G$ had two subgraphs $G_1$ and $G_2$ such that 
$G_1 \cup G_2=G$ and $V(G_1) \cap V(G_2)=\{v\}$.  Since $r(G)=d(G)=m$, there is a vertex
$y$ such that $d(v,y)=m$.  Suppose $y \in V(G_1)$.  Letting $z$ be any neighbour of $x$ in $G_2$, we have that $d(y,z)>m$, which contradicts $d(G)=m$.  
Now let $a$ and $b$ be two vertices such that $d(a,b)=m$.  By 2-connectivity, there is a cycle $C$ that contains both $a$ and $b$. We are done since $|C| \geq 2m$, else $d(x,y) <m$.  

I think if you choose $C$ as short as possible, you can do some re-routing to show that in fact $|C| \leq 2m+1$, but I haven't thought about this too much.