Yes.
For each finite group $Q$ I'll construct a finite group $H$ with $\mathrm{Out}(H)\simeq Q$, moreover $H$ will be constructed as a semidirect product $D\ltimes P$, with $P$ a $p$-group of exponent $p$ and nilpotency class $<p$, (with prime $p$ arbitrary chosen $>|Q|+1$) and $D$ abelian of order coprime to $p$ (actually $D$ being a power of a cyclic group of order $p-1$).
I'll use Lie algebras which are convenient tools to encode $p$-groups when $p$ is less than the nilpotency class, taking advantage of linear algebra.
In Lie algebras, we denote $[x_1,\dots,x_m]=[x_1,[x_2,\dots,[x_{m-1},x_m]\cdots]]$. Also I choose the convention to let permutations act on the left.
The base field is $K=\mathbf{F}_p$, $p$ prime. Fix $n\ge 1$. Let $\mathfrak{f}_n$ be the free Lie $K$-algebra on generators $(e_1,\dots,e_n)$. It admits a unique grading in $\mathbf{Z}^n$ such that $e_i$ has degree $E_i$, where $(E_i)$ is the canonical basis of $\mathbf{Z}^n$, it is called multi-grading. For instance, $[e_3,[e_1,e_3]]$ has multi-degree $(1,0,2,0,\dots,0)$.
Let $I$ be a finite-codimensional multi-graded ideal contained in $[\mathfrak{f}_n,\mathfrak{f}_n]$: so the quotient $\mathfrak{g}=\mathfrak{f}_n/I$ is naturally multi-graded. There is a natural action of ${K^*}^n$ on $\mathfrak{g}$: namely $(t_1,\dots,t_n)$ acts on $\mathfrak{g}_{(m_1,\dots,m_n)}$ by multiplication by $\prod_{i=1}^n t_i^{m_i}$. Let $D\subset\mathrm{Aut}(\mathfrak{g})$ be the image of this action. Also denote by $c$ the nilpotency class of $\mathfrak{g}$: we assume $p>c$.
Using that $p>c$, we endow, à la Malcev–Lazard, $\mathfrak{g}$ with the group law given by the Baker-Campbell-Hausdorff formula: $xy=x+y+\frac12[x,y]+\dots$. We thus view $\mathfrak{g}$ as both a Lie algebra and a group; we denote it as $G$ when endowed with the group law (but feel free to refer to the Lie algebra law in $G$); this is a $p$-group of exponent $p$ and nilpotency class $c<p$. Define $H=D\ltimes G$.
Every permutation $\sigma\in\mathfrak{S}_n$ induces an automorphism $u_\sigma$ of $\mathfrak{f}_n$, defined by $u_\sigma(e_i)=e_{\sigma(i)}$. Write $\Gamma_I=\Gamma_{\mathfrak{g}}=\{\sigma\in\mathfrak{S}_n:u_\sigma(I)=I\}$.
Proposition 1. The natural map $\Gamma_\mathfrak{g}\to\mathrm{Out}(H)$ is an isomorphism.
We need a lemma:
Lemma 2. Define $M$ as $\mathbf{F}_p^*\ltimes\mathbf{F}_p$. Then $\mathrm{Out}(M^n)$ is reduced to the symmetric group permuting the $n$ factors.
Proof. Let $f$ be an automorphism. This is a product of $n$ directly indecomposable center-free groups, hence its automorphism group permutes the $n$ (isomorphic) factors. Hence after composing with a permutation, we can suppose that $f$ preserves each factor. We are then reduced to checking that every automorphism of $\mathbf{F}_p^*\ltimes\mathbf{F}_p$ is inner. Indeed after composing by an inner automorphism, we can suppose that it maps the Hall subgroup $\mathbf{F}_p^*$ into itself. Then after composing with an inner automorphism in the given copy of $\mathbf{F}_p^*$, we can also suppose it acts as identity on $\mathbf{F}_p$. It easily follows that this is the identity. (Note for conciseness I used some slightly fancy results in this proof, but this lemma can be checked more elementarily.) $\Box$
Lemma 2'. [added May 8, 2024] Let $f$ be an automorphism of $H=D\ltimes G$ preserving $D$ such that the automorphism of $H/[G,G]\simeq M^n$ induced by $f$ is a plain permutation $\sigma$ of the $n$ factors. Then $\sigma\in\Gamma_I$ and $f$ permutes the $e_i$ as $\sigma$.
Proof. Write $f(e_i)=e_{\sigma(i)}+c_i$, where $c_i\in [\mathfrak{g},\mathfrak{g}]$. Write the graded decomposition $c_i=\sum_{k\in W_n}c_{i,k}$, where $W_n$ is the set of $n$-tuples of non-negative integers of sum $\ge 2$. For every $d\in D$ and every $i$, we have $f(de_id^{-1})=f(d)f(e_i)f(d)^{-1}$. Writing $d\cdot g$ for $dgd^{-1}$, this rewrites successively $$f(d\cdot e_i)=f(d)\cdot f(e_i)$$
(writing $d=(d_1,\dots,d_n)$): for $d_1,\dots,d_n\in\mathbf{F}_p^*$
$$f(d_ie_i)=(d_{\sigma^{-1}(1)},\dots)\cdot (e_{\sigma(i)}+\sum_{k\in\mathbf{Z}^n}c_{i,k}.$$ $$d_if(e_i)=d_ie_{\sigma(i)}+\sum_{k\in\mathbf{Z}^n}d_1^{k_{\sigma^{-1}(1)}}\dots d_n^{k_{\sigma^{-1}(n)}}c_{i,k}.$$
Taking the component of degree $k$ we get (for every $k,d,i$)
$$d_ic_{i,k}=d_1^{k_{\sigma^{-1}(1)}}\dots d_n^{k_{\sigma^{-1}(n)}}c_{i,k}.$$
Thus, for every $i,k$ such that $c_{i,k}\neq 0$, we obtain $d_i=d_1^{k_{\sigma^{-1}(1)}}\dots d_n^{k_{\sigma^{-1}(n)}}$ for all $d_1,\dots,d_n\in\mathbf{F}_p^*$. So, for such $i,k$, $d_i^{k_{\sigma^{-1}(i)}}-1)=1$ and $d_j^{k_{\sigma^{-1}(j)}}=1$ for all $d_1,\dots,d_n$. This means that, for each such $(i,k)$ $p-1$ divides $k_{\sigma^{-1}(i)}-1$ and divides $k_{\sigma^{-1}(j)}$ for all $j\neq i$. Since $c_{i,k}=0$ whenever component of $k$ is $\ge p-1$ (because the nilpotency length is $<p$ and each commutator involves at least two generators), this means that for every $i,k$ such that $c_{i,k}\neq 0$, we have $k_i=1$ and $k_j=0$ for all $j\neq i$. But since $k$ has total sum $\ge 2$, this is impossible. We deduce that $c=0$, i.e., $f$ permutes the generators.
Proof of the proposition.
Let $f$ be an automorphism of $H:D\ltimes G$. After composing with an inner automorphism, we can suppose that $f$ maps the Hall subgroup $D$ into itself.
Next, $f$ induces an automorphism of $H/[G,G]$, which can naturally be identified with $M^n$ of the previous lemma (recall that $I\subset [G,G]=[\mathfrak{g},\mathfrak{g}])$. Hence after composing with conjugation by some element of $D$, we can suppose that $f$ both preserves $D$ and acts on $H/[G,G]\simeq M^n$ by permuting the factors (without changing coordinates). [Edit May 8, 2024] By Lemma 2', $f$ permutes the generators $e_i$ and is induced by an element of $\Gamma_I$.[End edit — a few lines of the preceding proof now erased]$\Box$.
To conclude we have to prove:
Proposition 3. For every finite group $Q$ of order $n$ and prime $p>n+1$ there exist $n,p$ and $I$ finite-codimensional multi-graded ideal in $\mathfrak{f}_n(=\mathfrak{f}_n(\mathbf{F}_p)$, such that $\Gamma_I\simeq Q$. (And such that $\mathfrak{f}_n/I$ has nilpotency class $\le n+1$.)
This starts with the following, which provides for each finite group $Q$ a relation whose automorphism group is the group $L_Q\subset\mathfrak{S}(Q)$ of left translations of $Q$.
Lemma 4. Let $Q$ be a group, $n=|Q|$, and $q=(q_1,\dots,q_n)$ an injective $n$-tuple of $Q$. Define $X=Qq\subset Q^n$. Then $L_Q=\{\sigma\in\mathfrak{S}(Q):\sigma X=X\}$.
Proof. Clearly $L_Q$ preserves $X$. Conversely, if $\sigma$ preserves $X$, after composing with a left translation we can suppose that $\sigma(q_1)=q_1$, so $\sigma(q)\in\{q_1\}\times Q^{n-1}$; since $X\cap \{q_1\}\times Q^{n-1}=\{q\}$, we deduce $\sigma(q)=q$, which in turn implies $\sigma=\mathrm{Id}$. $\Box$.
Proof of the proposition.
Write $\mathfrak{f}_Q\simeq\mathfrak{f}_n$ as the free Lie algebra over the generating family $(e_q)_{q\in Q}$. It can be viewed as graded in $\mathbf{Z}^G$, with basis $(E_q)_{q\in Q}$. Write $E=\sum_q E_q$.
For $q\in Q^n$, define $\xi_q=[e_{q_n},e_{q_1},e_{q_2},\dots,e_{q_n}]$ (note that it is homogeneous of degree $E+E_{q_n}$; in particular $(\xi_h)_{h\in X}$ is linearly independent. Fix an injective $n$-tuple $q$ and define $X$ as in the proof of the lemma; for convenience suppose $q_n=1$.
Define $J$ as the $n$-dimensional subspace of $\mathfrak{f}_Q$ with basis $(\xi_h)_{h\in X}$.
Define $I=J\oplus \mathfrak{f}_Q^{n+2}$, where $\mathfrak{f}_Q^i$ is the $i$-th term in the lower central series. Hence $I$ is an ideal, and $\mathfrak{g}=\mathfrak{f}_Q/I$ is defined by killing all $i$-fold commutators for $i\ge n+1$ and certain particular $(n+1)$-fold commutators. (Since we assume $p>n+1$, we can view it as a group as previously.)
Claim. For $h=(h_1,\dots,h_n)\in Q^n$ with $h_{n-1}\neq h_n$, we have $\xi_h\in I$ if and only if $h\in X$.
By definition, $h\in X$ implies the condition. Now suppose that $h$ satisfies the condition. First, the condition $h_{n-1}\neq h_n$ ensures that $\xi_h\neq 0$; it is homogeneous in the multi-grading. If it belongs to $J$, its multi-degree is therefore some permute of $(2,1,\dots,1)$. This is the case if and only if the $h_i$ are pairwise distinct, so we now assume it; its degree is therefore equal to $E_{h_n}+E$. Now $J_{E+E_{h_n}}$ is 1-dimensional, and generated by $\xi_{h_nq}$. Hence $\xi_h$ is a scalar multiple of $\xi_{h_nq}$:
$$[e_{h_n},e_{h_1},\dots,e_{h_{n-1}},e_{h_n}]=\lambda
[e_{h_n},e_{h_nq_1},\dots,e_{h_nq_{n-1}},e_{h_n}].$$
The next lemma implies that $h_i=h_nq_i$ for all $i\in\{1,\dots,n-1\}$. So $h\in X$. The claim is proved.
The claim implies that for every permutation $\sigma$ of $Q$, if the automorphism $u_\sigma$ of $\mathfrak{f}_Q$ preserves $I$, then $\sigma$ has to preserve $X$, and hence (Lemma 4) $\sigma$ is a left translation of $Q$. This finishes the proof. $\Box$.
Lemma 5
Consider the free Lie algebra on $(e_1,\dots,e_n)$. If for some permutation $\sigma$ of $\{1,\dots,n-1\}$ and scalar $\lambda$ we have
$$[e_n,e_1,\dots,e_{n-1},e_n]=\lambda [e_n,e_{\sigma(1)},\dots,e_{\sigma(n-1)},e_n],$$
then $\sigma$ is the identity and $\lambda=1$.
Proof. Use the representation $f$ in $\mathfrak{gl}_n$ mapping $e_i$ to the elementary matrix $\mathcal{E}_{i-1,i}$ (consider indices modulo $n$). Then
$[e_n,e_1,\dots,e_{n-1},e_n]=[e_n,e_1,[e_2,\dots,e_n]]$ maps to $$[\mathcal{E}_{n-1,n},\mathcal{E}_{n,1},\mathcal{E}_{1,n}]=[\mathcal{E}_{n-1,n},\mathcal{E}_{n,n}-,\mathcal{E}_{1,1}]=\mathcal{E}_{n-1,n}.$$
Let by contradiction $j$ be maximal such that $\sigma(j)\neq j$; note that $2\le j\le n-1$ and $n\ge 3$. Then $[e_n,e_{\sigma(1)},\dots,e_{\sigma(n-1)},e_n]=[e_n,e_{\sigma(1)},\dots,e_{\sigma(j)},[e_{j+1},\dots,e_n]]$ maps to
$$w=[\mathcal{E}_{n,n-1},\mathcal{E}_{\sigma(1)-1,\sigma(1)},\dots,\mathcal{E}_{\sigma(j)-1,\sigma(j)},\mathcal{E}_{j,n}],$$
which cannot be zero. Hence $[\mathcal{E}_{\sigma(j)-1,\sigma(j)},\mathcal{E}_{j,n}]\neq 0$. Since $\sigma(j)<j$, this implies $\sigma(j)-1=n$ (modulo $n$), that is, $\sigma(j)=1$. So
$$w=-[\mathcal{E}_{n,n-1},\mathcal{E}_{\sigma(1)-1,\sigma(1)},\dots,\mathcal{E}_{\sigma(j-1)-1,\sigma(j-1)},\mathcal{E}_{j,1}].$$
In turn, $[\mathcal{E}_{\sigma(j-1)-1,\sigma(j-1)},\mathcal{E}_{j,1}]$, using that $\sigma(j-1)\neq 1$, implies $\sigma(j-1)=j$, and so on, we deduce that $\sigma$ is the cycle $j\mapsto j+1$ (modulo $n-1$). Eventually we obtain
$$w=[\mathcal{E}_{n-1,1},\mathcal{E}_{1,2},\mathcal{E}_{2,1}]=\mathcal{E}_{n-1,1}.$$
So $\mathcal{E}_{n-1,n}=\lambda \mathcal{E}_{n-1,1}$, contradiction.
(Note on Lemma 5: one has $[e_1,e_2,e_1,e_2]=[e_2,e_1,e_1,e_2]$, but this and its obvious consequences are probably the only identities between Lie monomials in the free Lie algebra, beyond the ones obtained from skew-symmetry in the last two variables.)
Note on the result: the resulting group $H$ roughly has size $|Q|^{|Q|}$, which is probably not optimal.
In Proposition 4, $I$ is strictly contained in $\mathfrak{f}_Q^{n+1}$, as soon as $|Q|\ge 3$, so the nilpotency class of $G$ is then equal to $n+1$. (For $|Q|=1$, choosing $p\ge 3$ outputs $H$ as the group $M=M_p$ which has trivial Out, and $G$ is abelian then; for $|Q|=2$, this outputs a group $H$ of order $(p-1)^2.p^3$ for chosen prime $p\ge 5$, and $G$ has nilpotency length $2$. For $|Q|=3$ it outputs a group $H$ of order $(p-1)^3.p^{29}$ for $p\ge 5$ which is already quite big.) To improve the bounds in explicit cases by running this method, one should describe $Q$ a permutation group of a set $Y$ that can be described as stabilizer of some $\ell$-ary relation $R$ contained in the set of pairwise distinct $\ell$-tuples of $Y$, for $\ell$ as small as possible.
