Are there recursive sets $X$ with Property A that contain infinitely many incompressible strings?

I assume that $X$ is a set of binary strings and the binary strings are coded by natural numbers in this way: consider this enumeration of all strings:

$\lambda$, 0,1,00,01,10,11,000,.....

(in which, at first we have the string with length zero, then all strings of length one in alphabetical orde and so on).Now code the $n$ th member of this enumeration by the number $n$.

Now we define the set $X$ as $\{w~:~ |w|=2^i~ for ~some ~i\in \mathbb{N}\}$ ($|w|$ denotes the length of the string $w$).

It can easily checked that $X$ satisfies the property $\textsf{A}$. It is also well-known that we have incompressible strings of any length (because the number of Turing machines of length<$n$ is at most $2^n-1$ but we have $2^n$ string of length $n$). Therefore $X$ contains infinitely many incompressible strings.

I'm assuming you call `incompressible' integers whose binary representation $x$ satisfy $K(x)>|x|-c$ for a fixed $c$?

If that's the case, the answer is no. If $X$ is a computable set of integers and some integer $k$ of representation $x$, with $|x|=N$, belongs to $X$, then $K(x|N) \leq \log |X^{\leq 2^{N+1}}| + c'$ for a fixed $c'$. (Indeed one can describe $x$ by just giving its position as a string of length $N$ inside $X$). But by your assumption on $X$, $\log |X^{\leq 2^{N+1}}| - N$ tends to $-\infty$, so almost all strings in $X$ satisfy, say, $K(x\mid|x|) < |x| - 3c$. You can then remove the the condition $|x|$ and get $K(x) < |x| - 2c$ by a usual padding technique (see the book by Li and Vitanyi, I can elaborate if this is important).