What follows is a proof that the conjecture is true for sufficiently large $n$.
Proof: We proceed by contradiction.
$n$ odd: If $n=2k+1$, then we may color the board as a checkerboard. Note that odd squares will always be the same color, and even squares will always be the same color. Since the knight alternates colors on each jump, and there is only one even prime, the result follows.
$n$ even: If $n=2k$, we similarly color the odd squares white, and the even squares black. Ignore the prime 2, so that all primes we are dealing with are odd. This means that we must always jump over two columns, and up or down 1. (Going up two rows and over 1 column would lead us to an even square which is non-prime)
Consider the set $\mathcal{P}_x$ to be the set primes less then $x$ where we have thrown out all pairs of primes $p,p+2\in \mathcal{P}$ and $p,p+4\in \mathcal{P}$. By using the Selberg sieve results, we know that are $\ll \frac{x}{\log^2 x}$ such pairs, and so it follows that $$|\mathcal{P}_x| \sim \frac{x}{\log x}$$ as $x\rightarrow \infty$. Now, choose $x\leq n$ so that all the elements of $\mathcal{P}_x$ lie in the first row. By the condition that we have no pairs of the form $p,p+2$, or $p,p+2$ in our set, and by the fact the knight must jump to a prime the next row, we see that among the integers $[n,n+x]$ we must have at least $|\mathcal{P}_x|$ primes. Baker, Harman and Pintz showed that $$\pi(x+x^{0.525})\sim \frac{x^{0.525}}{\log x},$$ (we can use a weaker result then this) and by taking $x=n^{0.6}$ we see that as $n\rightarrow \infty$ we must have $$\frac{n^{0.6}}{\log n}\sim \frac{n^{0.6}}{0.6\log n}$$ which is false.