do we at least have $Y_n=\Theta(\frac{\log n}{n})$ almost surely?
The answer to this is no. It is not even true that $Y_n=\Theta(\frac{h(n)}{n})$ almost surely (a.s.) for any (deterministic or random) function $h$ such that $h(n)\to\infty$ (as $n\to\infty$). Indeed, suppose the contrary: that $\liminf_n nY_n=\infty$ a.s. Then, by the Fatou lemma, $$\infty=E\infty\le\liminf_n EnY_n=\liminf_n n\frac1{n+1}=1,$$ a contradiction.
Remark 1: If $Y_n\sim f(n)$ a.s. for some deterministic positive function $f$, then, reasoning similarly to the above, we conclude that $f(n)=O(1/n)$. Indeed, otherwise $nf(n)\to\infty$ for $n=n_k$ and $k\to\infty$, where $(n_k)$ is some strictly increasing sequence of integers. So, by the Fatou lemma, $$1=E\lim\frac{Y_n}{f(n)}\le\liminf_n E\frac{Y_n}{f(n)}=\liminf_n \frac1{nf(n)}=0,$$ a contradiction.
Remark 2: It follows that, if $Y_n\sim f(n)$ a.s. for some deterministic positive function $f$, then without loss of generality $f$ is nonincreasing. Indeed, let $$g(n):=\min(f(1),\dots,f(n)).$$ Then $0<g(n)\le f(n)$ and $g(n)=f(k_n)$ for some integer $k_n$ between $1$ and $n$. Also, it follows from Remark 1 that $g(n)=O(1/n)=o(1)$. So, $k_n\to\infty$ and hence $$1\leftarrow\frac{Y_n}{f(n)}\le\frac{Y_n}{g(n)}\le\frac{Y_{k_n}}{f(k_n)}\to1,$$ so that $Y_n\sim g(n)$, and $g$ is a deterministic positive function.