Before an attempt to answer I'd like to add a few comments, some because they are simultaneously very interesting and hard to find in literature, some of them because they relate to the question. You decide which is which. :)

A. Niven's solution applies to polynomials $f(x)=\Sigma_{n=0}^N a_nx^n$, as you pointed out. Not all polynomial expressions have this form; for example, $xi+ix-1$ has no roots, where $i$ is a square root of $-1$.

B. Niven's proof that $f(x)$ has roots can be extended to $f(x)=g_N(x)+\Sigma_{n=0}^{N-1} h_n(x)$, where $g_N(x)$ is a **monomial** of degree $N$ and $h_n$ are polynomials of degree $n$. One just has to rephrase Gauss's proof for complex numbers, that relies on degrees of maps $S^1\to S^1$ (contraction of non-zero complex numbers to $U^1$) to degrees of maps $S^3\to S^3$ (contraction of non-zero quaternions to $Sp^1$). The picture that arises their is very interesting IMO.

C. Here's an interesting trick. Suppose that $q=xi+yj+zk+t$, where $i,j,k$ are the usual unit quaternions. Then you can recover $x,y,z,t$ as linear expressions of $q$:

$t = (q-iqi-jqj-kqk)/4$

$x = (q-iqi+jqj+kqk)/4$

$y = (q+iqi-jqj+kqk)/4$

$z = (q+iqi+jqj-kqk)/4$

This shows that any affine real algebraic variety in $R^4$ can be written up as the zero set of several quaternionic polynomials: just plug the above expressions over $q$ into the algebraic expression over the four real variables.

D. Notice that the above expressions for $x,y,z,t$ are rational. Therefore you can plug them into the equation for a rational variety: if $V\subset Q^4$ is the zero set of several equations in four variables $x,y,z,t\in Q$ then you just plug the above expressions and you get the same variety as the zero set of the corresponding equations in a single quaternionic variable $q\in L$. Here we identify $L$ with $Q^4$ in the obvious way.

E. Now, let's approach your question, let's restrict ourselves to $f(x)=\Sigma_{n=0}^N a_nx^n$, where $a_n\in L$. One of the things Niven showed that the zero sets of such polynomials consist of isolated points and 2-D spheres. The typical example for a sphere would be $f(x)=(x-a)^2+b$, where $a\in H$ is the center and $b\in R_+$ is the square of the radius.

F. So, in order to answer your question it's sufficient to ask ourselves: is it possible to embed $S^2\to R^3$ of imaginary quaternions so that (1) the center and the squared radius would be rational and (2) the intersection with $Q^3$ would be empty?

G. There is a result that the integers of the form $4n(8m+7)$ cannot be expressed as sums of three squares. So let's place the center of $S^2$ in the origin and pick radius 28. Then we have $f(x)=x^2+28$ that has no roots in $L$.

H. The above answers part (1) of the question, but obviously not part (2). I don't know off-hand how to extend this train of thought to finite extensions of $Q$.