Digital root of twin prime semiprimes

Your conjectured fact is true.

Of any $3$ consecutive integers, one is divisible by $3$. If $2$ of the integers are a pair of twin primes neither of which is $3$, then neither of the "end" numbers is divisible by $3$. So the "middle" number must be divisible by $3$.

So the twin primes are $3k-1$ and $3k+1$ for some integer $k$. It follows that their product is $9k^2-1=9k^2-9+8$. We conclude that the remainder when $(3k-1)(3k+1)$ is divided by $9$ is $8$.

But the remainder when a number $n$ is divided by $9$ is the same as the remainder when the sum of the decimal digits of $n$ is divided by $9$. Thus the digital root of our product is $8$.

The digital root of $n$ repeats with period $9$, viz. $1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, \ldots$ so the digital roots of a twin prime pair must be $(1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8), (7, 9)$, or $(8, 1)$. However, given that the digital root of a prime number greater than 3 cannot be $3, 6$ or $9$, that leaves $(2, 4), (5, 7)$, or $(8, 1)$. (adapted from a posting by David Radcliffe to a LinkedIn Number Theory group discussion)

And here's a link to a proof that contextualizes the digital root sequencing of twin primes: http://www.primesdemystified.com/twinprimesdigitalrootproof