Executive summary: If you look at the whole Hilbert scheme associated to a given polynomial, the locus of points corresponding to nonsingular (which I take to mean smooth) subschemes can sometimes be very small in terms of dimension and number of irreducible components. So in this sense, most subschemes are singular.
Details: The Hilbert scheme $\operatorname{Hilb}^P_{\mathbf{P}^n}$ associated to a given Hilbert polynomial $P$ is connected (a theorem of Hartshorne), but in general it has many irreducible components, each with its own generic point. Thus there are several different "generic" closed subschemes with the same Hilbert polynomial, each a member of a different family.
The locus of points in the Hilbert scheme corresponding to smooth (=nonsingular) subschemes of $\mathbf{P}^n$ is a Zariski open subset, which implies that it is Zariski dense in the union of the components that it meets, but there are often other components of the Hilbert scheme all of whose points correspond to singular subschemes.
Because the Hilbert scheme need not have a single generic point, one might ask: How many of these generic points parametrize singular subschemes, and what are the dimensions of the corresponding components of the Hilbert scheme?
As a case study, consider the Hilbert scheme $H_{d,n}$ of $d$ points in $\mathbf{P}^n$, i.e., the case where $P$ is the constant polynomial $d$. Points of $H_{d,n}$ over a field $k$ correspond to $0$-dimensional subschemes $X \subseteq \mathbf{P}^n$ of length $d$, or in other words, such that $\dim_k \Gamma(X,\mathcal{O}_X) = d$. Each smooth $X$ with this Hilbert polynomial is a disjoint union of $d$ distinct points. These smooth $X$'s correspond to points of an irreducible subscheme of $H_{d,n}$, and the closure of this irreducible subscheme is a $dn$-dimensional irreducible component $R_{d,n}$ of $H_{d,n}$. Sometimes $H_{d,n}=R_{d,n}$, which means that every $X$ is smoothable. But for each fixed $n \ge 3$, Iarrobino observed that $\dim H_{d,n}$ grows much faster than $\dim R_{d,n}$ as $d \to \infty$. (He proved this by writing down large families of $0$-dimensional subschemes, like $\operatorname{Spec} (k[x_1,\ldots,x_n]/\mathfrak{m}^r)/V$, where $\mathfrak{m}=(x_1,\ldots,x_n)$ and $V$ ranges over subspaces of a fixed dimension in $\mathfrak{m}^{r-1}/\mathfrak{m}^r$.) This shows that $H_{d,n}$ is not irreducible for such $d$ and $n$, and that the ``bad'' components all of whose points parametrize singular subschemes can have much larger dimension than the one component in which a dense open subset of points parametrize smooth subschemes. With a little more work, one can show that the number of irreducible components of $H_{d,n}$ can be arbitrarily large (and as already remarked, the components themselves can have larger dimension than $R_{n,d}$). So in this sense, one could say that for $n \ge 3$, most $0$-dimensional subschemes in $\mathbf{P}^n$ are singular.
For more details about $H_{d,n}$, including explicit examples of nonsmoothable $0$-dimensional schemes, see the following articles and the references cited therein:
The moduli space of commutative algebras of finite rank
Hilbert schemes of 8 points
(Warning: my notation $H_{d,n}$ is different from the notation of those articles.)
Best Answer
When you say "deformation functor", you have to be careful to specify exactly which functor you are thinking about. There are two relevant deformation functors at play here.
One is the functor $D$ of abstract deformations of $Y$. If $A$ is an Artin local $k$-algebra, then $D(A)$ is the set of isomorphism classes of flat families $Y_A$ over $A$ whose central fiber is isomorphic to $Y$.
The other functor $D'$ is the functor of deformations of $Y$ inside $\mathbb{P}^n_k$. Here, $D'(A)$ is the set of isomorphism classes of flat families $Y_A$ together with an embedding $Y_A \rightarrow \mathbb{P}^n_k \times Spec A$ such that the central fiber is $Y \rightarrow \mathbb{P}^n_k$ (not isomorphic to $Y$, but exactly $Y$, as we are talking about subschemes of $\mathbb{P}^n_k$).
The functor $D$ does not have any direct relationship to the Hilbert scheme, but the functor D' certainly does.
Given any scheme $X$ and a point $x \in X$, the local ring $\mathcal{O}_x$ defines a functor $F: (Art)_k \rightarrow (Sets)$ by setting $F(A) = Hom(Spec A, Spec \mathcal{O}_x)$. We say that $F$ is pro-represented by $\mathcal{O}_x$.
If you take the local ring $\mathcal{O}_{[Y]}$ of the Hilbert scheme (which you probably know to be representable by an honest scheme by a theorem of Grothendieck) at the point $[Y]$, then the functor it pro-represents is none other than $D'$.
Being pro-representable is stronger than having a versal deformation space; it means that the functor has a universal deformation space, which is the formal completion of the point whose local ring is doing the pro-representing.
So the (uni)versal deformation space of $D'$ is the formal completion of the point $[Y]$ in the Hilbert scheme. As for the versal deformation space of $D$, I am not really sure what it looks like, but there is a morphism $D' \rightarrow D$ of deformation functors given by forgetting the embedding of $Y_A \rightarrow \mathbb{P}^n_k \times Spec A$.
In general, I think of a versal deformation space as an infinitesimal object; it is only keeping track of what happens when you deform $Y$ a little bit. The Hilbert scheme, on the other hand, is global; it is keeping track of all subschemes with the same Hilbert polynomial as $Y$, including ones which might be far away from $Y$ (if $Y$ was a point, for instance). Thus, you would expect the versal deformation space of the appropriate functor to map into the Hilbert scheme, rather than the other way around.