We can prove something stronger, that there is no path at all. (That is removing the Hamiltonian condition) 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 if we are on a white square, we must jump to a white square. If we are on the first row, this tells us that the knights next move must be a jump to the second row, either two to the left, or two to the right.
Idea: We will use the above to show that we need at least the same number of primes in the second row as in the first, but this leads to a contradiction since the density of the primes decreases as we go to infinity. In particular, the second row's prime density will be much smaller then the first rows, implying the result.
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. Assuming the path exists, we must have a prime in the second row within jumping distance for each prime in the first row. (The knight cannot jump from first row to third) By the condition that we have no pairs of the form $p,p+2$, or $p,p+2$ in our set, we see that among the integers $[n,n+x]$ we must have at least $|\mathcal{P}_x|$ primes. (The condition regarding prime pairs makes sure that no prime in the second row is used as the jumping point for two distinct primes in the first row.) Baker, Harman and Pintz showed that $$\pi(n+n^{0.525})-\pi(n)\sim \frac{n^{0.525}}{\log n},$$ (we can use a weaker result then this) so taking $x=n^{0.525}$ we see there are $\frac{n^{0.525}}{\log n}$ primes in the interval $[n,n+x]$. However, we had bounded this below by $|\mathcal{P}_x|\sim \frac{n^{0.525}}{\log n^{0.525}}$, and this gives the asymptotic inequality $$\frac{n^{0.525}}{\log n} \gtrsim \frac{1}{0.525} \frac{n^{0.525}}{\log n},$$ which is evidently false.
Remark: We do not need a result as strong as Baker, Harman and Pintz, we need only $$\pi(x+x^{\theta})-\pi(x)\sim \frac{x^\theta}{\log x}$$ for some $\theta<1$.
Remark: If we wanted only to prove the conjecture for Hamiltonian paths, we did not need to waste time removing the twin primes pairs for fear of double counting, since a Hamiltonian path implies the inequality immediately.