I think the answer is YES. Yes, I thought of an alleged demonstration like this:
How, by hypothesis $\{T(t)\: ; \: t\geq 0\}$ is the $C_0$ semigroups,in $X$, (where $X=(X,|\cdot|)$ is a Banach space) then, by definition, we have $$\lim_{t\rightarrow 0}T(t)x=x,\: \forall \: x\in X,$$ that is, given any $\epsilon>0$, exists $\delta>0$ such that if $t\geq0$ and $|t|<\delta$ then $$ |T(t)x-x|<\epsilon,\: \forall\:x\in X. $$
In particular, for all $x\in X$ such that $|x|=1$, we have $$ |T(t)x-x|<\epsilon \: \Rightarrow \: \sup_{\stackrel{\scriptstyle x\in X}{|x|=1}} |T(t)x-x| <\epsilon $$ but, $$ \|T(t)-I\|=\sup_{\stackrel{\scriptstyle x\in X}{|x|=1}} |T(t)x-x| <\epsilon ,\: \forall \: \epsilon>0,\;|t|<\delta ,$$ where $I$ is identity operator, in $X$. Therefore, $\displaystyle \lim_{t\rightarrow 0} \|T(t)-I\|=0;$ that is, $\{T(t)\: ; \: t\geq 0\}$ is a uniformly continuous semigroup.
Is that correct?
You have operator norm continuity of the semigroup at $0$ iff the generator is bounded. Any unbounded generator results only in strong continuity at $0$. In the case of a uniformly continuous semigroup $T(t)$, the generator $A$ is a bounded operator and $T(t)=e^{tA}=\sum_{n=0}^{\infty}\frac{t^n}{n!}A^n$.
As an example, let $H=\ell^2$, and let $\{ e_n \}_{n=0}^{\infty}$ be the standard orthonormal basis. Then $$ T(t)x = \sum_{n=1}^{\infty}e^{-nt}\langle x,e_n\rangle e_n,\;\;\; t \ge 0, $$ is a $C_0$ semigroup with $\|T(t)\| \le 1$ for all $t \ge 0$. But it is not uniformly continuous at $t=0$ because $$ \|T(t)e_n-e_n\|=(1-e^{-nt}), $$ which cannot be made uniformly small for all $n=0,1,2,3,\cdots$, regardless of how small $t > 0$ is taken to be.