$\newcommand\si\sigma$Note that 
$$\si(T)=\frac1T\,\int_0^T dt\,s(t).$$
Take any $L>\limsup s(T)$ and then take any real $A>0$ such that $s(t)\le L$ for all real $t>A$. Then 
$$\limsup\si(T)\le\limsup\frac1T\,\int_0^A dt\,s(t)+\limsup\frac1T\,\int_A^T dt\,s(t)\le0+L=L,$$
for any $L>\limsup s(T)$. 

So, the answer to your first question is yes. 

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The answer to your second question is no. E.g., suppose that $s(t)=t\sin^2 t$ for real $t\ge0$. Then $\si(T)=\frac{1}{8} \left(2 T^2-2 T \sin (2 T)-\cos (2 T)+1\right)\to\infty$ but $s(T)\not\to\infty$ as $T\to\infty$.