>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 $h(n)\to\infty$ a.s. (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. $\quad\Box$

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Let us now show that $Y_n\not\sim f(n)$ a.s. for any deterministic positive function $f$. 

To prove this, suppose the contrary: that $Y_n\sim f(n)$ a.s. for some deterministic positive function $f$. Take any positive sequence $(a_n)$ such that $\sum_n a_n=\infty$. Then 
$$\sum_n P(X_n<a_n/2)=\sum_n a_n/2=\infty.$$
So, by the Borel--Cantelli lemma, events $\{X_n<a_n/2\}$ a.s. occur infinitely often (i.o.), that is, for infinitely many values of $n$. Therefore and because $\{X_n<a_n/2\}\subseteq\{Y_n<a_n/2\}$, we see that events $\{Y_n<a_n/2\}$ a.s. occur i.o. Recalling now the assumption that $Y_n\sim f(n)$ a.s., we conclude that $f(n)\le a_n$ i.o. In particular, $f(n)\le \frac1{n\ln n}=o(1/n)$ i.o. 

So, 
$$Z_n:=nY_n\sim nf(n)=o(1) \tag{1}\label{1}$$
a.s. for $n=n_k$ and $k\to\infty$, where $(n_k)$ is a strictly increasing deterministic sequence of positive integers. Also, $EZ_n^2<2$. So, for $n$ as above,
$$\begin{aligned}
1\leftarrow EZ_n&=EZ_n\,1(Z_n<4)+EZ_n\,1(Z_n\ge4) \\ 
&\le EZ_n\,1(Z_n<4)+EZ_n^2/4 \\ 
&<EZ_n\,1(Z_n<4)+1/2\to1/2  
\end{aligned}$$
by \eqref{1} and dominated convergence. So, we have a contradiction. $\quad\Box$