I thought I should expand a bit on my earlier answer.
As I see it, there are two questions here: do generalized functions
make sense on any Hilbert space and if not is there an analogue of
a classical Riesz representation theorem which is valid here?
To answer the first question, the first thing we need to rule out
is the $\delta$ function as a functional on $L^{2}$ I would argue
that the big issue here is that $L^{2}$ functions do not exist pointwise--you
can change their value on any set of measure zero without changing
the function. Because of this, there is no hope of making sense of
$\delta$ over the Hilbert space you are used to.
The natural question then becomes what space we should work on. I
would argue that the most important property of generalized functions
is that they are all 'smooth' in the distributional sense (i.e.
infinitely differentiable). If this is the property that you want
them to have, then you need at least that level of regularity on the
test functions you apply the generalized functions to (since by duality
we apply all the derivatives to the test function to define the distributional
derivative). This is essentially the trade-off one usually sees when
defining objects through duality: the more structure you put on the
test objects, the less structure you need on your dual objects to
get nice properties. This natural path leads one to the theory of
distributions, which are the continuous linear functionals on the
space $C_{0}^{\infty}$ with a particularly nasty (not even metrizible) but entirely natural topology on it.
For a sequence of test functions to converge in $C_{0}^{\infty}$
we need two things: they need to live in a single common compact set
(so no infinitely expanding supports) and all of the partial derivatives
have to converge in the infinity norm, i.e. $\|\partial^{\alpha}\varphi_{n}-\partial^{\alpha}\varphi\|_{\infty}\to0$
for all multiindices $\alpha.$ Continuity of a functional linear
functional $T$ in the dual means that if $\varphi_{n}\to\varphi$
in this topology, then $T\varphi_{n}\to T\varphi.$ As a comment,
this is not good enough to specify the topology since we are not working
on a metric space. To prove that this space is not metrizible, we
would actually have to go through the nitty gritty details of its
construction. It's not too terrible if you have a background in topological
vector spaces, but it's not edifying if you don't so I won't. The
important property for our purposes is that there is no metric space
with the same topology as the one we gave to $C_{0}^{\infty}.$ In
particular, it does not admit a Hilbert space structure. Now, Hilbert
spaces are all self-dual (this is the Riesz theorem you are used to),
which means that $C_{0}^{\infty*}$also does not admit a Hilbert space
structure. The answer to the first question is therefore no. Generalized
functions do not make sense over a Hilbert space.
Before going into the second question, let me comment that there are
a lot of 'Riesz representation theorems.' All that these theorems
have in common is that over many nice spaces there is a natural class
of continuous linear functionals (on a Hilbert space, it is given
by the inner product; on $L^{p}$ it is given by multiplication by
an $L^{q}$ function and justified by Holder's inequality; on $C_{0}$
it is given by integration against a Borel measure). The common trait
of Riesz-type theorems is that they say that all functionals belong
to this natural class of functionals. The obvious class in our case
is integration against Borel measures (since our functions are still
continuous), but we cannot apply the classical Riesz-Markov theorem
since we changed the topology of the space we are working over. It
turns out the result is false in any case: the objects we are working
with have too much structure for measures to be able to distinguish
between all of them.
The fact that the $\delta$ function is represented as a measure when
viewed as a functional over $C_{0}^{\infty}$ is a manifestation of
the fact that all positive distributions (that is all continuous linear
functionals $T$ on $C_{0}^{\infty}$ with the property that if $f\geq0$
then $Tf\geq0$) are represented as measures.
For an example of why we cannot apply Riesz-Markov here, I do not believe that the generalized function given by the derivative of the Dirac delta is represented as a measure.