Yes, it follows that $P(2)$ is a composite integer. (Note: 1 is neither prime nor composite.)

Write $P(x) = a_0 + a_1 x + \cdots + a_n x^n$ with $a_i \in \{0,1\}$ and $a_n = 1$.
Let $\alpha$ be an arbitrary root of $P$. Since $P(\alpha) = 0$, $\alpha$ cannot be a positive real number.
%In addition, if $\alpha \neq 0$ then
%\begin{align*} %1 &< |a_{n-1} \alpha^{-1} + \cdots + a_0 \alpha^{-n}| \\ %&\leq |\alpha|^{-1} + \cdots + |\alpha|^{-n} %\end{align*}
%and so $|\alpha| < 2$.
%
In addition, if $\alpha \neq 0$ then
$$\begin{align*} |1 + a_{n-1} \alpha^{-1}| &= |a_{n-2} \alpha^{-2} + \cdots + a_0 \alpha^{-n}| \\ &\leq |\alpha|^{-2} + \cdots + |\alpha|^{-n}. \end{align*}$$
If $\alpha \neq 0$ and $\mathrm{Re}(\alpha) \geq 0$, then $\mathrm{Re}(1 + a_{n-1} \alpha^{-1}) \geq 1$
and
$$1 \leq |\alpha|^{-2} + \cdots + |\alpha|^{-n} < \frac{|\alpha|^{-2}}{1 - |\alpha|^{-1}};$$
this yields $|\alpha| < (1 + \sqrt{5})/2$.

By the same token, if $\alpha \neq 0$ then
$$|1 + a_{n-1} \alpha^{-1} + a_{n-2} \alpha^{-2}| \leq |\alpha|^{-3} + \cdots + |\alpha|^{-n}.$$
We deduce from this that $\mathrm{Re}(\alpha) \leq 3/2$ as follows.
\begin{itemize}
\item
There is nothing to check if $\mathrm{Re}(\alpha) \leq 0$.
\item
If the argument of $\alpha$ belongs to $[-\pi/4, \pi/4]$, then $\mathrm{Re}(\alpha^{-1}), \mathrm{Re}(\alpha^{-2}) \geq 0$, so
$$1 \leq |\alpha|^{-3} + \cdots + |\alpha|^{-n} < \frac{|\alpha|^{-3}}{1 - |\alpha|^{-1}}.$$
Hence $|\alpha|^{-1}$ is greater than the unique positive root of $x^3 + x - 1$, which
is greater than $2/3$.
\item
Otherwise, $\alpha$ has argument in $(-\pi/2,\pi/4) \cup (\pi/4,\pi/2)$,
so the bound $|\alpha| < (1 + \sqrt{5})/2$ implies that $\mathrm{Re}(\alpha) < (1 + \sqrt{5})/(2 \sqrt{2}) < 3/2$.
\end{itemize}

By hypothesis, there exists a factorization $P(x) = Q(x)R(x)$ into two nonconstant integer polynomials, which we may assume are monic.
$Q(x + 3/2)$ is a product of polynomials, each of the form $x - \alpha$ where $\alpha$ is a real root of $P$
or of the form
$$\begin{align*} &\left( x + \frac{3}{2} - \alpha\right) \left(x + \frac{3}{2} - \overline{\alpha} \right) \\ &\quad = x^2 + 2 \mathrm{Re}\left(\frac{3}{2} - \alpha\right) x + \left|\frac{3}{2} - \alpha \right|^2 \end{align*}$$
where $\alpha$ is a nonreal root of $P$. It follows that $Q(x+3/2)$ has positive coefficients;
comparing its values at $x=1/2$ and $x=-1/2$ yields $Q(2) > Q(1)$. We cannot have $Q(1) \leq 0$, as otherwise the intermediate value theorem would imply that $Q$ has a real root in $[1, \infty)$; hence $Q(1) \geq 1$ and so $Q(2) \geq 2$.
Similarly $R(2) \geq 2$, so $P(2) = Q(2) R(2)$ is composite.

\textbf{Remark.}
A theorem of Brillhart, Filaseta, and Odlyzko from 1981 states that if a prime $p$ is written as $\sum_i a_i b^i$ in any base $b \geq 2$, the polynomial $\sum_i a_i x^i$ is irreducible.
(The case $b=10$ is an older result of Cohn.)
The solution given above is taken from: Ram Murty, Prime numbers and irreducible polynomials, \textit{Amer. Math. Monthly} \textbf{109} (2002), 452--458). The final step is due to P\'olya and Szeg\H{o}.