The answer is yes.
By Dirichlet's theorem, there are infinitely many primes of the form $210k+1$ (this is a special case provable in elementary way). But then $210k+2,210k+4,210k+6,210k+8,210k+10$ are even, $210k+3,210k+9$ are divisible by $3$, $210k+5$ is divisible by $5$ and $210k+7$ is divisible by $7$. Hence $210k+1,...,210k+10$ form a block of 10 consecutive numbers, only one of which is prime.
Let me mention that this method can be extended to blocks of any integer length. The only property of $210$ which was used here is that it is divisible by all the primes up to $10$.
Another proof, based on djechlin's idea in a comment to different answer: for $n\geq 10$, let $p$ be the first prime greater than $n!+1$. Then the gap between $n!+1$ and $p$ contains only composite numbers, and there are at least 9 of them, since $2\mid n!+2,3\mid n!+3,...,10\mid n!+10$, so last 9 composites together with $p$ form a set like the one you asked for.
This gives infinitely many such sets, because we can choose $n$ arbitrarily large.