I have the following continuous-time matrix Riccati equation
$$A X + X A' - X b b' X + Q = 0$$
where $Q>0$, $A$ is a diagonal matrix with strictly negative eigenvalues and $b$ is a (column) vector.
Without knowing $b$, only that $\| b \|_2 \leq \beta$, can I find a diagonal matrix $Y$, with strictly positive eigenvalues, such that $Y\leq X$, (where $Y\leq X$ means that $v'Yv\leq v'Xv$ for any vector $v$)?
On
Okay, here is an attempt at a solution. As @Kwin-van-der-Veen says, there is a unique symmetric positive definite $X$ which satisfies the equation $AX+XA^{T}-Xbb^{T}X+Q=0$.
Now let $q_{min}:=min_{x}\left\{ x^{T}Qx:\;\left\Vert x\right\Vert =1\right\} $ and $a_{max}:=max_{x}\left\{ -x^{T}Ax:\;\left\Vert x\right\Vert =1\right\} $.
Pick $0<\delta<\frac{q_{min}}{3*max\left\{ a_{max},\beta\right\} }$, and let $\bar{A}=A-\delta bb^{T}$ and $\bar{Q}=Q+2\delta A-\delta^{2}bb^{T}$.
We can show that $X\geq\delta I$.
As $A$ is a diagonal matrix with strictly negative eigenvalues, it follows that $\bar{A}$ is a symmetric strictly negative definite matrix. It also follows that $\bar{Q}$ is positive semi-definite, as $x^{T}Qx+2\delta x^{T}Ax-\delta^{2}xbb^{T}x\geq0$ for any vector $x$ and equal to zero only when $x=0$.
From this, there is a unique symmetric positive definite matrix $Z$ which satisfies
$\bar{A}Z+Z\bar{A}'-Zbb'Z+\bar{Q}=0$
It follows that $X=Z+\delta I$ is the unique symmetric positive definite solution to $AX+XA'-Xbb'X+Q=0$ and $X\geq\delta I$.
Since $A$ only has negative eigenvalues then for any choice for $b$ the pair would be stabilizable. Combining this with $Q\geq0$ ensures that there is always a positive definite solution for $X$. This means that $X$ only has strictly positive eigenvalues. Now by setting every diagonal element of $Y$ to the smallest eigenvalue of $X$ then it should satisfy your requirement.