(I) The derivative of the function is $f'(x)=e^{x}-a$.
$1.$ When $a\leqslant 1$, $f'(x)=e^{x}-a\geqslant 0$ holds for any $x\geqslant 0$, so $f(x)$ is a monotonically increasing function on $(0,+\infty)$; Since $f(0)=0$, $f(x)\geqslant f(0)=0$ holds for any $x\geqslant 0$.
$2.$ When $a > 1$, let $f'(x)=0$, we get $x=\ln a > 0$. When $x\in(0,\ln a)$, $f'(x) 0$, $f(x)$ is monotonically increasing. If $f(x)\geqslant 0$ holds for any $x\geqslant 0$, then only need $f(x)\_{min}=f(\ln a)=e^{\ln a}-a\ln a-1=a-a\ln a-1\geqslant 0$. Let $g(a)=a-a\ln a-1 (a > 1)$, then $g'(a)=1-\ln a-1=-\ln a 1$.
In conclusion, $a\leqslant 1$.

(II) From (I), when $a=1$, $f(x)=e^{x}-x-1$, $f'(x)=e^{x}-1$. It is easy to get $f(x)\_{min}=f(0)=0$, i.e., $e^{x}\geqslant x+1$ holds for any $x\in R$. Taking $x=- \frac {3i-1}{3n}(i=1,2,…,n)$, we get $1- \frac {3i-1}{3n}\leqslant e^{- \frac {3i-1}{3n}}$, i.e., $(1- \frac {3i-1}{3n})^{n}\leqslant (e^{- \frac {3i-1}{3n}})^{n}\leqslant e^{( \frac {3i-1}{3n})^{n}}=e^{- \frac {3i-1}{3}}$. Summing up, we get $(1- \frac {2}{3n})^{n}+(1- \frac {5}{3n})^{n}+…+(1- \frac {3n-1}{3n})^{n}\leqslant e^{- \frac {2}{3}}+e^{- \frac {5}{3}}+…+e^{- \frac {3n-1}{3}}$. Therefore, $(3n-2)^{n}+(3n-5)^{n}+…+1^{n}\leqslant [e^{- \frac {2}{3}}+e^{- \frac {5}{3}}+…+e^{- \frac {3n-1}{3}}](3n)^{n}=e^{- \frac {2}{3}} \frac {1- \frac {1}{e^{n}}}{1- \frac {1}{e}}(3n)^{n} < \frac {e^{ \frac {1}{3}}}{e-1}(3n)^{n}$. So when $n\geqslant 2, n\in N$, it always holds that $1^{n}+4^{n}+7^{n}+…+(3n-2)^{n} < \boxed{\frac {e^{ \frac {1}{3}}}{e-1}(3n)^{n}}$.