Yes, every prime occurs as a prime factor of a highly composite number. Suppose towards a contradiction that it does not; then let $p_n$ be the least prime which does not occur as a factor. Note that this means a highly composite number cannot contain any primes higher than $p_n$ either, since switching out such a prime for $p_n$ gives a smaller number with the same number of divisors. Thus, this would mean that all highly composite numbers are of the form
$$
2^{k_1} \cdot 3^{k_2} \cdots p_{n-1}^{k_{n-1}}.
$$
Since there are infinitely many highly composite numbers, at least one of the values $k_i$ must be unbounded; hence, there is a prime $p_i$ and a highly composite number such that $p_n < p_i^{\lfloor k_i/3 \rfloor }$. But now note that removing $\lfloor k_i/3\rfloor$ copies of the prime $p_i$ from this number lowers the total number of divisors by less than $1/3$, while adding the prime $p_n$ doubles its number of divisors. This contradicts the assumption that this was the prime factorization of a highly composite number.
Thus, $p_n$ occurs in the prime factorization of a highly composite number.
Edit: Since this question has received so much attention, I feel like it might be worth explaining some of the intuition behind this answer. Recall, a highly composite number is a number which has more divisors than any number before it. Now, if
$$
p_1^{k_1} \cdots p_n^{k_n}
$$
is any prime factorization of a number, then that number will have
\begin{equation}
\qquad\qquad\qquad\qquad\qquad(1 + k_1) \cdots (1 + k_n)\qquad\qquad\qquad\qquad(1)
\end{equation}
divisors: namely, for every prime $p_i$ a divisor can include that prime $0, 1, \ldots, k_i$ times, which gives $1 + k_i$ options. Note that the primes don't really factor into this anymore; swapping out all copies of a prime with a prime that did not yet occur in the number leaves the $k_i$ and therefore the number of divisors unchanged.
Of course, which primes occur in the number does matter for the size of the number: higher primes give a higher number.
Thus, we think of constructing highly composite numbers as a sort of optimization problem: by multiplying by a new prime factor, we "buy" some new divisors, at the "expense" of making the number larger.
At first, we use low prime factors for that; multiplying by 2, 3, 5 is much "cheaper" than multiplying by 9999991. However, there is a law of diminishing returns. The first time you add a prime factor, you double the number of divisors: in formula (1), a factor $(1 + 0)$ is replaced by one of the form $(1 + 1)$. The second time you do it, a factor $(1 + 1)$ is replaced by $(1 + 2)$, providing only a multiplier of $\frac32$ to the number of divisors.
So, the more and more of the same prime factor you add, the less returns you get off that. If you keep it up long enough, higher primes become more "lucrative". Eventually, you will have exhausted so much of the usefulness of all primes below 9999991, that it becomes optimal to add the prime factor 9999991. The proof just makes this precise. With a little more fidgeting, you can prove that in fact every number, not just every prime, divides a highly composite number.
This limit doesn't exist; the proportion oscillates because the highest numbers dominate it. For instance, for $p_n\approx5\cdot10^k$, most numbers will be less than their reversed versions, simply because they start with a digit from $1$ to $4$, whereas for $p_n\approx10^k$ the proportion will be roughly $\frac12$.
Instead, you should consider the logarithmic density, which weights each range $\left[10^n,10^{n+1}\right]$ equally.
By the prime number theorem for arithmetic progressions, the primes are asymptotically equidistributed over the admissible residue classes mod $10^k$. Thus, $\frac14$ of them end in each of $1,3,7,9$, the ones ending in $1$ are equidistributed with respect to ending in $01,11,\ldots,91$, and so on. Thus, the reversed version of a prime is effectively a random number with the first digit uniformly randomly drawn from $1,3,7,9$ and the following digits uniformly randomly drawn from all digits. A number $x$ distributed in $\left[10^n,10^{n+1}\right]$ with density $\frac1{x\log10}$ thus has the following probability to be less than its reversed version:
\begin{eqnarray}
&&
\frac1{4\log10}\left(\int_1^2\frac{2-x}x\mathrm dx+\log3+\int_3^4\frac{4-x}x\mathrm dx\\+\log7+\int_7^8\frac{8-x}x\mathrm dx+\log9+\int_9^{10}\frac{10-x}x\mathrm dx\right)
\\
&=&
\frac{2\log2+\log7+8\log2-4\log3+24\log2-8\log7+\log9+9\log10-18\log3-4}{4\log10}
\\
&=&
\frac{43\log2-20\log3+9\log5-7\log7-4}{4\log10}
\\
&\approx&
0.5099\;.
\end{eqnarray}
This is the logarithmic density of the primes that are less than their reversed versions.
Best Answer
If a composite number is divisible by 3, the sum of its digits is also divisible by 3.
If a composite number is divisible by 11, the alternating sum of its digits is also divisible by 11.
Just two of those facts will give you a $\frac13+\frac1{11}=43\%$ better chance. You can further increase this number by starting adding up rarer cases. For example, if the number of digits in $p$ is a multiple of 3, then if $p = \overline{a_1a_2a_3a_4a_5a_6\ldots}$ is divisible by 7, then $(a_1+2a_2+a_3)+(a_4+2a_5+a_6)+...$ is also divisible by 7.