As Pavel says, any semisimple element lies in a maximal torus, where you can take any root. On the other hand, if $g$ is unipotent, then it is in the image of the exponential map, so you can make sense of $g^\lambda$ for any $\lambda \in \mathbb C$, so counterexamples must have nontrivial semisimple and unipotent parts.
Suppose $g=s.u=u.s$ is the Jordan decomposition of an element. Then $s^n$ and $u^n$ are semisimple and unipotent respectively, so they are the Jordan decomposition of $g^n$. Thus the existence of roots is compatible with Jordan decomposition.
Now take a semisimple element $s$ such that $Z_G(s)^0$ does not contain a central torus (so its center is finite) -- if $G$ is simply connected then in fact $Z_G(s)$ is connected, so I'll assume that. Now pick a regular unipotent element $u$ in $Z_G(s)$ and consider
$g =s.u$. I want to claim $g$ is a counterexample. Indeed suppose for each $n$ we have $h_n$ an $n$-th root, and $h=s_nu_n$ is its Jordan decomposition. Then $s_n^n =s$ and $u_n^n=u$, and both $s_n$ and $u_n$ lie in $Z_G(s)$. Then we see that $s_n$ centralizes $u$ for all $n$, but since $u$ is regular in $Z_G(s)$ and $s_n$ semisimple it follows that $s_n$ must lie in the centre of $Z_G(s)$. But then taking, say, $n$ equal to the order of that centre (which is finite) we get a contradiction, as $s_n^n$ must then be $1$.
Semisimple elements $s$ such that $Z_G(s)$ does not contain a central torus exist, but there are only finitely many conjugacy classes of them, as was essentially shown in the paper of Borel and de Siebenthal. In fact I think that paper establishes that there are $r+1$ such classes where $r$ is the rank of $G$, so this would give a negative answer for the exceptional groups also. I suspect that these might somehow be the only counterexamples?