I am grateful to Will Sawin and Fedor Petrov for the inspiration their answers provided me. Thanks also to Will Jagy for some bounds on the divisor function $\tau(n)$. Although I still need to show that $F(n) = \sum_{1 \lt d, d|n} \phi(d)/\log d \lt 1.95n/\log n$ holds for all $n \lt 10^{11}$, when I have that, I can use the argument below to show the inequality holds for all $n$. I will use $C$ and later set $C=1.95$.

Let's split the index set into divisors $d \leq d'=n^r$, and $d \gt d'$:
$$F(n) = \sum_{1\lt d\leq d', d|n} \frac{\phi(d)}{\log d} + \sum_{d' \lt d, d|n} \frac{\phi(d)}{\log d} \leq \frac{\tau(n)d'}{\log d'} + \frac{n-d'}{\log d'}.$$
The bound of the first sum comes from overestimating the number of terms by $\tau(n)$ and each term by $d'/\log d'$. The bound of the second sum arises from replacing the denominator $\log d$ by $\log d'$ and using a property of $\phi$. For $r=1/2$, and writing $\log d'$ as $(\log n)/2$, we get an upper bound of $(\tau(n)\sqrt{n} + n - \sqrt{n})2/\log n$. As $\tau(n) \leq \sqrt{3n}$, so $F(n)\lt (2+2\sqrt{3}) n/\log n$ for all $n\gt 1$.

We can remove a $\sqrt{3}$ from the constant by observing that $\tau(n)$ can be replaced by the sharper count $(\tau(n)-1)/2$, and since $\tau(n)$ is eventually bounded by $n^s$ for any real $s \gt 0$ and $n$ sufficiently large, we get that the sum of interest is eventually bounded by $2n/\log n$. We can do better.

Let's increase $r$ slightly. Rewriting the right hand side, we get $$F(n) \lt \frac{(\tau(n)-3)n^r +n}{r\log n}.$$ Recounting the divisors (and excluding $1$ and $n$) shows there are at most $\tau(n) - 2$ many which are greater than $1$ and at most $d' \lt n$. For fixed $r$ and increasing $n$, we get that eventually $F(n) \lt n(1+ \epsilon)/r\log n$. Even though this can eventually approach $1$ by increasing $r$ and $n$, we want to do better, especially as $\tau(n)n^r$ stays close to or above $n$ for some moderately large $n$ and $r\gt 1/2$.

As $r$ gets larger, the estimate for the upper sum shrinks and approaches the true value of this sum, while the bound on the lower sum grows like $\tau(n)n^r/r\log n$. I need a better estimate of the lower sum if I want $F(n) \lt Cn/\log n$ for all $n$, not just sufficiently large $n$.

I thus assume $F(m) \lt Cm/\log m$ for all $m$ less than a bound $N$, and then use an improved estimate to get $F(n) \lt Cn/\log n$ for $n \geq N$. I use (assuming all $d|n$) $$\sum_{1\lt d \leq d'} \frac{\phi(d)}{\log d} \leq \sum_{d \in S} \frac{Cd}{\log d} \leq \frac{|S|Cd'}{\log d'}.$$ Here $S$ is a maximal irredundant set of covering divisors $d$ with $d \leq d'$. So for any divisor $f \leq d'$ of $n$, there is at least one $d \in S$ with $f|d$, and no two members of $S$ divide one another. Then every term $\phi(f)/\log f$ is incorporated into some term $Cd/\log d$ for one or more $d \in S$.

The improvement comes from seeing that $|S|\leq h(n)=\tau(n)/(1+a_e)$, with $a_e$ the largest exponent occuring in the prime factorization of $n$. (Proof of this will appear elsewhere, say in a comment.) I encourage study of the growth of $h$-champions, which seem to be more sparsely distributed than highly composite numbers.

I now choose $d'$ according to a tunable parameter $k$, so that I guarantee $|S|Cd'/\log d' \leq n/k\log n$, or in terms of $r=\log d'/\log n$, so that $|S|Ck \leq rn^{1-r}$. Then when $h(n)C/r \leq n^{1-r}/k$ and $1/k + 1/r \leq C$, we get the desired inequality.

We have $h(n) \leq \tau(n)/2$ for all $n$. This seems to tip the balance.
When we pick $4/7$ or $5/9$ for $r$, and choose $k$ to be $4$ or $20/3$ respectively to get $(1/r + 1/k)=C=1.95$, computer simulation suggests $h(n)Ck \lt rn^{1-r}$ for all $n \gt 5$ times $10^8$. On the theory side, we compare $\tau(n)Ck/2r$ with $n^{1-r}$ using upper bounds on $\tau(n)$ like $1152(n/367567200)^{0.244651}$ to get that the bound holds for all $n\gt 1.3$ times $10^{10}$. So the bound $F(n) \lt 1.95n/\log n$ will be established for all $n$ once it is established for all the troublesome cases below $10^{11}$. Computer experimentation gives that $F(n)\log n/n$ achieves a maximum of less than $1.607$ when $n=60$ for integral $n \in [2,10^8)$, so I'm feeling pretty good that $C=1.95$ will work for all $n$.

Gerhard "Gentlemen, Start Your Processor Cores" Paseman, 2015.11.29