The proof of contractibility in the strong(=weak=compact-open) topology is very easy:

Identify $H$ with $L^2([0,1])$.

The path $\{u_t\}$ that connects any unitary $u$ to the identity element is then given by:

$$u_t:L^2([0,1])= L^2([0,t])\oplus L^2([t,1])\qquad\qquad\qquad\qquad\qquad\qquad\qquad$$ $$\qquad\qquad\qquad
\xrightarrow{\varphi_t u \varphi_t^{-1}\,\oplus\, id}
L^2([0,t])\oplus L^2([t,1])=L^2([0,1])
$$

were $\varphi_t:L^2([0,1])\to L^2([0,t])$ is the unitary given by
the obvious reparametrization.

To answer your question, I would say that it's a case-by-case analysis:

*Examples:*

For an infinite algebraic direct sum of Hilbert spaces, I can mimic the above argument and make it work: use $\bigoplus^\infty L^2([0,1])$ and do the same shrinking to $\bigoplus^\infty L^2([0,t])$.

For the $n$-th algebraic tensor power of a Hilbert space, I can also make it work.
Again, it's a matter of finding a 1-parameter family of embeddings of the pre-Hilbert space into itself that "shrinks it to zero". Think about $L^2([0,1]^n)$ shrinking to $L^2([0,t]^n)$ and note that it preserves the subspace given by the $n$-th algebraic tensor power of $L^2([0,1])$.

For exterior powers, it's the same as for tensor powers.

For your particular example of the algebraic Fock space, that same strategy also works, and so the unitary group is strong-contractible.
Think of $\mathcal F$ as given by by $\bigoplus_{n=0}^\infty\bigwedge^n (L^2([0,1]))$ and note that you can shrink it to $\bigoplus_{n=0}^\infty\bigwedge^n (L^2([0,t]))$. Once again, this shrinking procedure preserves the "algebraic" subspace of the Hilbert space $\mathcal F$.