Is ${F_{n}}^2 - 28$ always a composite number?
${F_n}^2 - 28$ indeed can not be a prime number for $n>5$.
Let's suppose that ${F_n}^2 - 28$ is a prime. In that case, $n$ must be even, this was shown (easily using congruences) in other answers, so I'll omit this part for space saving. I'll also use known identity $5{F_n}^2 = {L_n}^2 - 4$ valid for even $n$ ($L_n$ is $n$-th Lucas number). It follows that that:
$$5({F_n}^2-28) = 5{F_n}^2-140 = ({L_n}^2 - 4) - 140 = {L_n}^2-144 = (L_n-12)(L_n+12)$$
${F_n}^2 - 28$ could be prime only if one of $L_n - 12$ and $L_n + 12$ are $1$, $5$, $-1$, $-5$, and those are just few cases of small $n$. For all others, ${F_n}^2 - 28$ can not be a prime.
Possible generalization: Expressions ${F_n}^2 - (k^2 - 4)/5$ (k is any natural number) could be candidates for a similar proof. However, some other bits and peaces (like proving that $n$ must be even) must be valid too, and this is often not true. However, it looks that, for instance, ${F_n}^2 - 145$ satisfies all conditions, and also can be reduced to $({L_n}^2 - 4) - 725$ and finally to $(L_n-27)(L_n+27)$.
Outside such generalization, it seems that the original statement (or its slightly modified version related only to condition $n>5$) is valid for ${F_n}^2 - 13$, ${F_n}^2 - 31$, ${F_n}^2 - 45$, ${F_n}^2 - 58$, ${F_n}^2 - 78$, ${F_n}^2 - 85$, ${F_n}^2 - 91$, ${F_n}^2 - 115$, ${F_n}^2 - 133$, ${F_n}^2 - 142$, ${F_n}^2 - 154$, ${F_n}^2 - 175$, ${F_n}^2 - 211$, ${F_n}^2 - 217$. (warning: I checked then with the help of a computer, and only for $n<1000$; please see comments for an interesting example involving $n>1000$)
Now, for the sake of curiosity, let us take a look at one of these cases. Case ${F_n}^2 - 85$ can be proven using congruences for $2$, $3$, $17$, and $107$ only. This means that one of $2$, $3$, $17$, and $107$ is always a divisor of ${F_n}^2 - 85$. Original case ${F_n}^2 - 28$ is fundamentally different, and, if I may say, "more beautiful", in the sense that it can have (and often has) only huge divisors, like in these examples:
$${F_{20}}^2 - 28 = 3023\times15139$$ $${F_{172}}^2 - 28 = 10348333\times53505724471\times315619755257\times10455376853041\times82024860865049\times1040059595540327$$
For users of Mathematica, this code:
Select[Table[Fibonacci[n], {n, 1, 1000}], PrimeQ[#*# + 3] &]
returns primes of the form ${F_n}^2+3$, for $1\le n\le 1000$. In this case there are $6$ such primes:
{
2, 8, 3524578, 27777890035288, 2011595611835338993891308327102733537615455242513357158345612749706882\ 9146295425939723629305572732574726246290673965789878845363842331040064\ 16432124798818261534841714338, 2949592466076064248964701302014885591673737506156850406413751530665307\ 5810241060939483954895520932111023343610904846943097162533007651451709\ 723277579925520157875345780869307228929160}
You're correct: $F_n^2-28$ is never a prime for $n\geq 6$, i.e., all $n$ where $F_n^2-28\gt 0$; arguably it's prime for $n=4$ ($F_4^2-28=-19$) and $n=5$ ($F_5^2-28=-3$).
First of all, note that if $3\not\mid F_n$, then $F_n^2\equiv 1\equiv 28\pmod 3$, so $3\mid F_n^2-28$. We can therefore assume in what follows that $3\mid F_n$ and therefore that $4\mid n$.
Now, (defining $G_n=F_n^2-28$ for convenience in what follows:) numeric evidence suggests the conjecture that for all $n$ divisible by $4$ there's some $a$ with $G_n=5a^2+24a = a\times(5a+24)$. We can prove this (and thus provide a factorization for all $n$) as follows:
Suppose that $G_n=5a^2+24a$; in other words, $5a^2+24a-G_n=0$. The (positive) solution to this quadratic equation is $a=\frac1{10}\left(-24+\sqrt{24^2+20G_n}\right)$. We'll first show that this quantity is rational, and then that it's integral. Note that $24^2+20G_n=24^2+20(F_n^2-28)=16+20F_n^2$, so $\sqrt{24^2+20G_n}=2\sqrt{4+5F_n^2}$. But since $n$ is even, we have $4+5F_n^2=L_n^2$ where $L_n$ is the $n$th Lucas number (a sort of conjugate to the Fibonacci numbers, satisfying the same recurrence). Clearly $2L_n-24$ is divisible by $2$; its divisibility by $5$ is equivalent to saying that $2L_n\equiv -1\bmod 5$ or that $L_n=2\bmod 5$. But this follows since, as noted at the start, we're in the case $4\mid n$ (and because the period of the Lucas numbers mod $5$ is $4$). This implies the integrality of $a$, which in turn implies the desired factorization of $G_n$.
The recurrence for the squares of the Fibonacci numbers is $$ x_n=2x_{n-1}+2x_{n-2}-x_{n-3} $$ Thus a repeat of a subsequence of length $3$ implies a repeat of the entire sequence.
mod $2$, we get $$ \overbrace{0,1,1},\overbrace{0,1,1},\dots $$ Thus, for $n\equiv0\pmod3$, we have that $F_n^2-28\equiv0\pmod2$
mod $3$, we get $$ \overbrace{0,1,1},1,\overbrace{0,1,1},\dots $$ Thus, for $n\not\equiv0\pmod4$, we have $F_n^2-28\equiv0\pmod3$
mod $7$, we get $$ \overbrace{0,1,1},4,2,4,1,1,\overbrace{0,1,1},\dots $$ Thus, for $n\equiv0\pmod8$, we have that $F_n^2-28\equiv0\pmod7$
The only thing not covered above is $n\equiv4\pmod{24}$ and $n\equiv20\pmod{24}$.
Recall that $F_n=\frac{\alpha^n-\beta^n}{\alpha-\beta}$ where $\alpha$ and $\beta$ are the roots of $x^2-x-1=0$. Note that $(\alpha-\beta)^2=5$ and $\alpha\beta=-1$. Using these relations, it is not difficult to obtain $$ \begin{align} 5(F_{2k}^2-28) &=5F_{2k}^2-140\\ &=\alpha^{4k}+\beta^{4k}-142\\ &=\left(\alpha^{2k}+\beta^{2k}\right)^2-144\\ &=\left(L_{2k}-12\right)\left(L_{2k}+12\right) \end{align} $$ where $L_n=\alpha^n+\beta^n$ is a Lucas Number. In an argument similar to those above, it can be shown that $L_{2k}-12\equiv0\pmod5$ for even $k$ and $L_{2k}+12\equiv0\pmod5$ for odd $k$.
This covers the case of all even values of $n$ for $n\ge6$ since $L_6=18$ and $L_n$ is monotonically increasing after that.
The cases above cover all $n\ge6$, which are the cases for which $F_n^2-28\ge0$. Specifically:
If $n$ is odd, $F_n^2-28=3\cdot\frac{F_n^2-28}3$
If $n=0\pmod4$, $F_n^2-28=\frac{L_n-12}{5}(L_n+12)$
If $n=2\pmod4$, $F_n^2-28=9\cdot\frac{L_n-12}3\frac{L_n+12}{15}$
The argument above can be simplified by ignoring the cases mod $2$ and mod $7$ since only the case mod $3$ and the Lucas number argument is needed.
I see that the Lucas number identity has already been noted in another answer while I've worked on this answer, but I will include it for completeness.