The answer is yes. Let $$J = \left( {\begin{array}{*{20}c}
\hfill {J_1} & \hfill {} & \hfill {} & \hfill {} \\
\hfill {} & \hfill {J_2} & \hfill {} & \hfill {} \\
\hfill {} & \hfill {} & \hfill \ddots & \hfill {} \\
\hfill {} & \hfill {} & \hfill {} & \hfill {J_n} \\
\end{array}} \right)$$ be your Jordan block matrix. Each Jordan block $J_i$ has a singular value decomposition $J_i = U_i \Sigma _i V_i ^* $, so we can write out a singular value decomposition for $J$ as $$J = \left( {\begin{array}{*{20}c} \hfill {U_1} & \hfill {} & \hfill {} & \hfill {} \\
\hfill {} & \hfill {U_2} & \hfill {} & \hfill {} \\
\hfill {} & \hfill {} & \hfill \ddots & \hfill {} \\
\hfill {} & \hfill {} & \hfill {} & \hfill {U_n} \\
\end{array}} \right)\left( {\begin{array}{*{20}c}
\hfill {\Sigma _1} & \hfill {} & \hfill {} & \hfill {} \\
\hfill {} & \hfill {\Sigma _2} & \hfill {} & \hfill {} \\
\hfill {} & \hfill {} & \hfill \ddots & \hfill {} \\
\hfill {} & \hfill {} & \hfill {} & \hfill {\Sigma _n} \\
\end{array}} \right)\left( {\begin{array}{*{20}c}
\hfill {V_1^* } & \hfill {} & \hfill {} & \hfill {} \\
\hfill {} & \hfill {V_2^* } & \hfill {} & \hfill {} \\
\hfill {} & \hfill {} & \hfill \ddots & \hfill {} \\
\hfill {} & \hfill {} & \hfill {} & \hfill {V_n^* } \\
\end{array}} \right).$$
The fact that we can express the singular value decomposition of $J$ in compatible block form with the Jordan blocks $J_i$ is essentially the only observation needed to answer your question.
$\left\| J \right\|_2 = \sigma _{\max}(J)$, so assume $\sigma_{max}(J)$ belongs to $\Sigma_k$ (it has to belong to some $\Sigma_i$ for $1\le i\le n$). Then, since $J_k=U_k\Sigma_k V_k^*$, we have that $$\|J\|_2=\sigma_{max}(J)=\|\Sigma_k\|_2=\|J_k\|_2=\max_{i=1,\cdots,n}\|J_i\|_2.$$
Note that this argument does not involve anything specific about the Jordan block structure. It remains valid provided $J$ has any diagonal block structure.