This not an answer.
As a try, I used the Taylor expansion of $\log(1+t)$ (which converges for $t=1$) and wrote
$$\log(1+\sin^2(x))=\sum_{n=1}^\infty (-1)^{n+1}\frac{\sin^{2n}(x)} n$$ and, so, we face the problem of $$I_n=\int_0^\infty e^{-x} \sin^{2n}(x)\,dx=\frac{i \, (-1)^n\, n\, \Gamma \left(\frac{i}{2}-n\right)\, \Gamma (2
n)}{4^n\,\Gamma \left(n+1+\frac{i}{2}\right)}$$ making
$$\int_0^\infty e^{-x}\log(1+\sin^2(x))\,dx=-i\sum_{n=1}^\infty \frac{ \Gamma \left(\frac{i}{2}-n\right)\, \Gamma (2
n)}{4^n\,\Gamma \left(n+1+\frac{i}{2}\right)}$$
Let
$$a_n=-i\frac{ \Gamma \left(\frac{i}{2}-n\right)\, \Gamma (2
n)}{4^n\,\Gamma \left(n+1+\frac{i}{2}\right)}$$ which gives
$$a_{n+1}=-\frac{2 n (2 n+1)}{4 n (n+2)+5}\,a_n \qquad \text{with} \qquad a_1=\frac 25$$ This makes the summation easy to compute but the convergence is quite slow as shown in the table below for the partial sums
$$S_p=\sum_{n=1}^{10^p} a_n$$
$$\left(
\begin{array}{cc}
p & S_p \\
0 & 0.4000000000 \\
1 & 0.3002972944 \\
2 & 0.3058023962 \\
3 & 0.3059876518 \\
4 & 0.3059935444 \\
5 & 0.3059937309 \\
6 & 0.3059937368
\end{array}
\right)$$
while the "exact" value should be $0.3059937370$.
To give an idea about the numbers of terms to be added for a given number of significant figures, a quick-and-dirty regression gives
$$\log_{10} \left(\left|a_{10^p}\right|\right) \sim -0.41-1.5 p$$
Edit
Looking at metamorphy's solution and at Mariusz Iwaniuk's answer, I feel a bit ridiculous with the so many terms I have been using here. In fact, the exact result can be obtained much faster using Euler's transformation.
Concerning the summation given in metamorphy's answer, it converges extremely fast. Considering the partial sums
$$T_k=-4\sum_{n=1}^k \frac{ \left(3-2 \sqrt{2}\right)^n}{(4 n^2+1)n}$$
$$\left(
\begin{array}{cc}
k & T_k \\
1 & -0.1372583002 \\
2 & -0.1407215063 \\
3 & -0.1409035111 \\
4 & -0.1409168427 \\
5 & -0.1409180203 \\
6 & -0.1409181376 \\
7 & -0.1409181503 \\
8 & -0.1409181517 \\
9 & -0.1409181519
\end{array}
\right)$$ and, using a CAS,
$$\sum_{n=1}^{\infty}\frac{r^n}{n}\frac{2a}{n^2+a^2}=-\frac{r^{-i a} B_r(1+i a,0)+r^{i a} B_r(1-i a,0)+2 \log (1-r)}{a}$$
Update
After Mariusz Iwaniuk's answer, I considered the case of
$$J_k=\int_0^\infty e^{-x}\log(1+\sin^{2k}(x))\,dx$$ using the same procedure as above. The result for $k=1$ having been given, here are a few other (not simplified to show some patterns)
$$J_2=\frac{24}{85} \,
_5F_4\left(1,1,\frac{5}{4},\frac{6}{4},\frac{7}{4};\frac{3}{2}-\frac{i}{4},\frac
{3}{2}+\frac{i}{4},2-\frac{i}{4},2+\frac{i}{4};-1\right)$$
$$J_3=\frac{144}{629} \,
_7F_6\left(1,1,\frac{7}{6},\frac{8}{6},\frac{9}{6},\frac{10}{6},\frac{11}{6};\frac{4}{3}-\frac{i}{6},\frac{4}{3}+\frac{i}{6},\frac{5}{3}-\frac{i}{6},\frac{5}{3}+
\frac{i}{6},2-\frac{i}{6},2+\frac{i}{6};-1\right)$$ The next is already too long to fit in a line : so if will described as
$$J_4=\frac{8064}{40885} \,
_9F_8\left(1,1,\color{red}{\text{#}};\color{green}{\text{@}};-1\right)$$ where
$$\color{red}{\text{#}}=\frac{9}{8},\frac{10}{8},\frac{11}{8},\frac{12}{8},\frac{13}{8},\frac{14}{8},\frac{15}{8}$$
$$\color{green}{\text{@}}=\frac{5}{4}-\frac{i}{8},\frac{5}{4}+\frac{i}{8},\frac{3}{2
}-\frac{i}{8},\frac{3}{2}+\frac{i}{8},\frac{7}{4}-\frac{i}{8},\frac{7}{4}+\frac{
i}{8},2-\frac{i}{8},2+\frac{i}{8}$$
