The answer is no.

**Update.** Here is an explicit example (which also preserves the $\mathbb Q$-span of the basis). Let $V=\ell_2$ over $\mathbb R$ and $(e_j)$ the standard basis. Define $T:V\to V$ by
$$
\begin{aligned}
(Tx)_1 &= x_1+3x_2+x_3 \\
(Tx)_2 &= 3x_1+10x_2+6x_3+x_4 \\
(Tx)_n &= x_{n-2}+6x_{n-1}+11x_n+6x_{n+1}+x_{n+2}, \qquad n>2 .
\end{aligned}
$$
Let $\alpha$ be the root of $p(x)=x^2+3x+1$ such that $|\alpha|<1$. Then the vector $v\in\ell_2$ defined by $v_n=\alpha^n$ satisfies $Tv=0$, so it is an eigenvector for $\lambda=0$. On the other hand, no nonzero rational vector $w$ satisfies $Tw=0$ because the relation $Tw=0$ implies the recurrence relation
$$
w_{n+4}=-6w_{n+3}-11w_{n+2}-6w_{n+1}-w_n ,
$$
which does not have rational solutions tending to 0.

**Below is the original answer.**

Let $V=\ell_2$ over $\mathbb R$ and $(e_j)$ the standard basis, so that $\langle x,e_j\rangle=x_j$ for every $x\in\ell_2$. Let $v\in V$ be a unit vector with irrational ratios of coordinates (for definiteness, take a geometric progression like $(c\pi^{-n})_{n\in\mathbb N}$). Then there exist a continuous self-adjoint $T:V\to V$ with rational coordinates such that $\ker T=\langle v\rangle$. So we have $\lambda=0$ but no $\lambda$-eigenvector is rational.

I represent $T$ by its (infinite) matrix $(t_{ij})_{i,j\in\mathbb N}$, $t_{ij}=\langle Te_i,e_j\rangle$. This matrix should be symmetric, and continuity of the resulting operator should be taken care of.

Begin with a self-adjoint $A$ such that $\ker A=\langle v\rangle$ with irrational components $a_{ij}$. For example, let $A$ be the projection to the orthogonal complement of $v$, so $a_{ij}=\delta_{ij}-v_iv_j$. I am going to approximate the matrix $(a_{ij})$ by a rational matrix $(t_{ij})$ such that the kernel stays the same.

Since $Av=0$, each row of our matrix is orthogonal to $v$. Approximate the first row $a_i=(a_{1i})$ by a rational vector $t_1=(t_{1i})$ such that $\langle t_1,v\rangle=0$ and $|t_{1i}-a_{1i}|$ is bounded by a rapidly decaying geometric progression (see below for details). Replace the first row and the first column of the matrix by this approximation. Then adjust the diagonal elements $a_{ii}$ so that the rows remain orthogonal to $v$, namely change $a_{ii}$ to $a_{ii}-(a_{1i}-t_{1i})v_1v_i^{-1}$. Recall that $v_i$ is a not-so-fast decaying geometric progression, so this adjustment is a small change.

Now remove the first row and the first column from the matrix, and the first component from $v$. Apply the same procedure to the truncated data: namely approximate the first remaining row and the first remaining column by a rational vector whose scalar product with $v$ is the same as before, then adjust the diagonal elements so that the scalar products of all rows with $v$ are preserved. Repeat ad infinitum. Note that every element of the original matrix is changed only finitely many times, and $v$ belongs to the kernel of the matrix after each step.

The approximations are controlled and the adjustments are bounded in terms of approximations, so the sum $\sum (a_{ij}-t_{ij})^2$ can be made arbitrarily small. This implies that $T$ is continuous and $\|T-A\|$ is small (say, less than 1/2), so we have $Tv=0$ and $\|Tw\|\ge\frac12\|w\|$ for all $w$ orthogonal to $w$. Therefore $\ker T=\langle v\rangle$. However $v$ cannot be rescaled to a rational vector.

It remains to show that every vector $w\in\ell_2$ can be approximated by a rational vector $w'$ such that $\langle w',v\rangle=\langle w,v\rangle$ and $w'_i-w_i$ is bounded by a small, fast decaying progression. Let $w'_1=w_1+\varepsilon_1$ be a rational approximation of $w_1$, then let $w'_2=w_2-\varepsilon_1v_1v_2^{-1}+\varepsilon_2$ be a rational approximation of $w_2-\varepsilon_1v_1v_2^{-1}$, then let $w'_3=w_3-\varepsilon_2v_2v_3^{-1}+\varepsilon_3$ be a rational approximation of $w_3-\varepsilon_2v_2v_3^{-1}$, and so on. The $\varepsilon_i$ at each step can be chosen arbitrarily small, and the resulting vector $w'$ satisfies $\langle w'-w,v\rangle=0$.